Question 1011 Mark
ABCD is a parallelogram, M is the mid-point of BD and BM bisects $\angle\text{B}.$ Then, $\angle\text{AMB}=$
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M.C.Q
Question 1021 Mark
Answer
View full question & answer→- 35º, 100º, 105º, 120º
Solution:
given: $\angle\text{A}+ \angle\text{C}=140^\circ$
and $\angle\text{A}:\angle\text{C} = 1:3$
and $\angle\text{B}:\angle\text{D} = 5:6$
$⇒ \angle\text{A}= \frac{1}{4} \times 140^\circ - 35^\circ$
$⇒ \angle\text{C}= \frac{3}{4} \times 140^\circ - 105^\circ$
Now according to angle sum property of quadrilateral
$\angle\text{A} +\angle\text{B}+ \angle\text{C}+ \angle\text{D} = 360^\circ$
$\Rightarrow 35^\circ+ \angle\text{B}+ 105^\circ +\angle\text{D} = 360^\circ$
$\Rightarrow \angle\text{B}+ \angle\text{D} = 360^\circ - 140^\circ = 220^\circ$
$\Rightarrow 5\text{x} + 6\text{x} = 220^\circ$
$\Rightarrow\text{x}=20^\circ$
So, $\angle\text{B} = 5 \times 20^\circ = 100^\circ$
and $\angle\text{D} = 6 \times 20^\circ = 120^\circ$
Question 1031 Mark
The bisectors of any two adjacent angles of a parallelogram intersect at:
Question 1041 Mark
In each of the questions one question is followed by two statements I and II. Choose the correct option.
Is quadrilateral ABCD a parallelogram?
Is quadrilateral ABCD a parallelogram?
- Diagonals AC and BD bisect each other.
- Diagonals AC and BD are equal.
Answer
View full question & answer→- If the question can be answered by one of the given statements alone and not by the other.
Solution:
Here, as we know that if the diagonals of a quadrilateral bisects each other, then it is a parallelogram.
But as per II, if the diagonals of a quadrilateral are equal, then it is not necessarily a parallelogram which is not true. Thus, II does not give the answer.
So the question can be answered by the one of the given statement alone and not by the other.
Question 1051 Mark
The angle between two altitudes of a Parallelogram through the vertex of an obtuse angle of the Parallelogram of 60º. Find the angles of the Parallelogram?
Answer
View full question & answer→- 120º, 60º, 120º, 60º
Solution:
Let ABCD be a parallelogram and AP and CQ are the altitudes drawn from vertex A on sides DC and BC.
In quadrilateral APCQ, sum of the all angles= 360º
So, $60^\circ + 90^\circ + \angle\text{C} + 90^\circ = 360^\circ$
$\angle\text{C} = 360^\circ - 240^\circ = 120^\circ$
$\angle\text{C} + \angle\text{B} = 180^\circ$ (co-interior angles)
$\angle\text{B} = 180^\circ - 120^\circ = 60^\circ$
In parallelogram, opposite angles are equal.
So, $\angle\text{A} = \angle\text{C} = 120^\circ$ and $\angle\text{B} = \angle\text{D} = 60^\circ$
Question 1061 Mark
What is the length of PQ in a trapezium ABCD in which $\text{AB || DC}$ and P and Q are mid-points on AD and BC respectively?
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Question 1071 Mark
Answer
View full question & answer→- x = 55º and y = 110º
Solution:
ABCO is a rectangle The diagonals of a rectangle are congruent and bisect each other. Therefore, in
$\triangle\text{AOB},$ we have:
OA = OB
$\angle\text{OAB} = \angle\text{OBA} = 35^\circ$
$\text{x} = 90^\circ - 35^\circ = 55^\circ$ and $\angle\text{AOB} = 180^\circ - (35^\circ + 35^\circ) = 110^\circ$
$\text{y} = \angle\text{AOB} = 110^\circ$ [Vertically opposite angles]
Hence, x = 55° and y = 110°
Question 1081 Mark
Write the correct answer in the following:
ABCD is a rhombus such that $\angle\text{ACB}=40^\circ.$ then $\angle\text{ADB}$ is:
View full question & answer→ABCD is a rhombus such that $\angle\text{ACB}=40^\circ.$ then $\angle\text{ADB}$ is:
Question 1091 Mark
Answer
View full question & answer→- 15cm
Solution:
In triangle ABC, P, Q and R are the midpoints.
By midpoint theorem, PQ is parallel to BC and $\text{PQ} = \frac{1}{2}$ of BC
QR is parallel to AB and $\text{QR} = \frac{1}{2}$ of AB
PR is parallel to AC and $\text{PR} = \frac{1}{2}$ of AC.
So, perimeter of triangle $\text{PQR} = \text{PQ} + \text{QR} + \text{PR} = \frac{1}{2}$ of $\text{(AB + BC + AC)}$
$= \frac{1}{2}$ of $(10 + 8 + 12) =\frac{1}{2} $ of $30=15.$
Question 1101 Mark
In $\triangle\text{ABC},$ E is the mid-point of median AD such that BE produced meets AC at F. If $\text{AC} = 10.5\text{cm},$ then AF = ?
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Question 1111 Mark
Answer
View full question & answer→- 40º
Solution:
$40^\circ, \ \angle\text{C} = 60^\circ$ as opposite angles of a parallelogram are equal and $\angle\text{COB} = 40^\circ$ angle sum property of a triangle. [In $\triangle\text{COB}, \ \angle\text{C} + \angle\text{COB} + \angle\text{OBC} = 180^\circ$]
Question 1121 Mark
Given Rectangle ABCD and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. If length of a diagonal of Rectangle is 8cm, then the quadrilateral PQRS is a:
Answer
View full question & answer→- Rhombus with each side 4cm.
Solution:
A quadrilateral formed by joining midpoints of the sides of the rectangle is a Rhombus.
In Rhombus, all sides are equal.
In triangle ABC, P and Q are midpoints of the sides AB and BC respectively. By midpoint theorem, PQ is parallel to AC and PQ is half of the AC.
Let diagonal AC = 8cm. So, PQ = 4cm.
Therefore, PQRS is a rhombus which all sides equal to 4cm.
Question 1131 Mark
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a:
Answer
View full question & answer→- Rectangle.
Solution:
The figure formed by joining the mid points of the adjacent sides of a rhombus is a rectangle.
Question 1141 Mark
Opposite angles of a Quadrilateral ABCD are equal. If AB = 4cm, find the length of CD.
Answer
View full question & answer→- 4cm
Solution:
A quadrilateral with both pair of opposite angles equal is a parallelogram.
In a parallelogram, opposite sides are equal.
So, AB = CD = 4cm
Question 1151 Mark
Answer
View full question & answer→- x = 30º and y = 95º
Solution:
Given, ABCD is a parallelogram.
So,
$\angle\text{A} = \angle\text{C}$ (Opposite angles of parallelogram are equal in size)
⇒ 3x − 20 = x + 40
⇒ 3x − x = 40 + 20
⇒ 2x = 60
⇒ x = 30°
Thus, $\angle\text{A} = 3 × 30 − 20 = 90 − 20 = 70^\circ$
Now, $\angle\text{A} + \angle\text{B} = 180^\circ$ (Sum of interior angles of parallelogram is 180º)
$⇒ 70^\circ + \angle\text{B} = 180^\circ$
$⇒ \angle\text{B} = 180^\circ − 70^\circ$
$⇒\angle\text{B} = 110^\circ$
$⇒ \text{y} + 15 = 110^\circ$
$⇒ \text{y} = 95^\circ$
Hence, x = 30° and y = 95°
Question 1161 Mark
In a rhombus ABCD, if $\angle\text{ACB}=40^\circ,$ then $\angle\text{ADB}=$
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Question 1171 Mark
Diagonals necessarily bisect opposite angles in a:
Question 1181 Mark
In Quadrilateral ABCD, $\angle\text{A} = 110^\circ, \ \angle\text{B} = 75^\circ$ and $\angle\text{C} = 35^\circ.$ Find $\angle\text{D}=\ ?$
Answer
View full question & answer→- 140º
Solution:
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property of quadrilateral).
$110 + 75^\circ + 35^\circ + \angle\text{D} = 360^\circ$
$\angle\text{D} = 360^\circ - 220^\circ = 140^\circ$
Question 1191 Mark
The angle between two altitudes of a Parallelogram through the vertex of an obtuse angle of the Parallelogram of 60º. Find the angles of the Parallelogram.
Answer
View full question & answer→- 120º, 60º, 120º, 60º
Solution:
Let ABCD be a parallelogram and AP and CQ are the altitudes drawn from vertex A on sides DC and BC.
In quadrilateral APCQ, sum of the all angles = 360º
So, $60^\circ + 90^\circ + \angle\text{C} + 90^\circ = 360^\circ$
$\angle\text{C} = 360^\circ - 240^\circ = 120^\circ$
$\angle\text{C} + \angle\text{B} = 180^\circ$ (co-interior angles)
$\angle\text{B} = 180^\circ - 120^\circ = 60^\circ$
In parallelogram, opposite angles are equal.
So, $\angle\text{A} = \angle\text{C} = 120^\circ$ and $\angle\text{B} = \angle\text{D} = 60^\circ$
Question 1201 Mark
ABCD is a Trapezium in which $\text{AB || DC}$ and $\angle\text{A}=\angle\text{B} = 45^\circ.$ Find $\angle\text{C}$ and $\angle\text{D}$ of the Trapezium.
Answer
View full question & answer→- 135º, 135º
Solution:
AB is parallel to DC.
$\angle\text{A} + \angle\text{D} = 180^\circ$ (co-interior angle)
$\angle\text{D} = 180^\circ - 45^\circ = 135^\circ$
Similarly by following same argument, $\angle\text{C} = 135^\circ.$
Question 1211 Mark
The diagonal AC and BD of quadrilateral ABCD are equal and are perpendicular bisector of each other then quadrilateral ABCD is a:
Answer
View full question & answer→- Square
Solution:
The triangles formed by the perpendicular bisectors are congruent to each other by SAS congruence. so the sides are equal hence it is a square.
Question 1221 Mark
Angles of a quadrilateral are in the ratio 3 : 4 : 4 : 7. Find all the angles of the quadrilateral.
Answer
View full question & answer→- 60º, 80º, 80º, 140º
Solution:
Let ABCD be a quadrilateral with $\angle\text{A} = 3\text{x},\ \angle\text{B} = 4\text{x} ,\ \angle\text{C} = 4\text{x}$ and $\angle\text{D} = 7\text{x}$
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property)
$3\text{x} + 4\text{x} + 4\text{x} + 7\text{x} = 360^\circ$
$18\text{x} = 360^\circ$
$\text{x} = 20^\circ$
$\angle\text{A} = 3 (20^\circ) = 60^\circ$
$\angle\text{B} = \angle\text{C} = 4 (20^\circ) = 80^\circ$
$\angle\text{D} = 7 (20^\circ) = 140^\circ$
Question 1231 Mark
D and E are the mid-points of the sides AB and AC res. Of $\triangle\text{ABC}.$ DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is:
Answer
View full question & answer→- $\text{DE = EF}$
Solution:
If DE = EF then triangle AED becomes congruent to triangle CEF by SSS congruence rule.
By $\text{CPCT}, \ \angle\text{ECF} = \angle\text{EAD}$ which forms a pair of alternate angles.
Which proves that AD is parallel to CF.
Question 1241 Mark
In the given figure, $\text{ABCD}$ is a Rectangle. Find the values of $x$ and $y$?
Answer
View full question & answer→Given $\angle\text{AOB}=110^\circ$
$\Rightarrow \angle\text{DOC}=110^\circ$ vertically opposite angles
$\triangle\text{DOC}$ we have:
$DO = OC$
Now, $\angle\text{ODC} = \angle\text{OCD} = \text{y}$
now in $\triangle\text{ODC}$
$y + y + 110^\circ = 180^\circ ($angle sum property of triangle$)$
$\Rightarrow 2y = 180^\circ - 110^\circ = 70^\circ$
$\Rightarrow y = 35^\circ$
Also$, x = 90^\circ – y$
$x = 90^\circ – 35^\circ = 55^\circ$
Hence,$ x = 55^\circ$ and $y = 35^\circ$
$\Rightarrow \angle\text{DOC}=110^\circ$ vertically opposite angles
$\triangle\text{DOC}$ we have:
$DO = OC$
Now, $\angle\text{ODC} = \angle\text{OCD} = \text{y}$
now in $\triangle\text{ODC}$
$y + y + 110^\circ = 180^\circ ($angle sum property of triangle$)$
$\Rightarrow 2y = 180^\circ - 110^\circ = 70^\circ$
$\Rightarrow y = 35^\circ$
Also$, x = 90^\circ – y$
$x = 90^\circ – 35^\circ = 55^\circ$
Hence,$ x = 55^\circ$ and $y = 35^\circ$
Question 1251 Mark
Answer
View full question & answer→- 45º
Solution:
It is given in the question that,
In parallelogram ABCD: $\angle\text{BAD} = 75^\circ, \ \angle\text{CBD} = 60^\circ$
Now, $\angle\text{DAB} = \angle\text{DCB} = 75^\circ$ (Opposite angles)
Also, in triangle DBC we know that sum of angles of a triangle is 180º
$\angle\text{DBC} + \angle\text{BDC} + \angle\text{DCB} = 180^\circ$
$60^\circ + \angle\text{BDC} + 75^\circ= 180^\circ$
$135^\circ + \angle\text{BDC} = 180^\circ$
$\angle\text{BDC} = 180^\circ – 135^\circ$
$\angle\text{BDC} = 45^\circ$
Hence, 45º is correct.
Question 1261 Mark
If one angle of a parallelogram is 24º less than twice the smallest angle, then the measure of the largest angle of the parallelogram is:
Question 1271 Mark
Answer
View full question & answer→- 7.2cm
Solution:
BC = 4 × 1.8 = 7.2cm.
Question 1281 Mark
A quadrilateral ABCD is a parallelogram if:
Answer
View full question & answer→- $\angle\text{A} = 60^\circ, \ \angle\text{C} = 60^\circ,\ \angle\text{B} = 120^\circ$
Solution:
$\angle\text{A} = 60^\circ, \ \angle\text{C} = 60^\circ,\ \angle\text{B} = 120^\circ$
Opposite angles are equal and sum of adjacent angles are supplementary.
Question 1291 Mark
In a quadrilateral ABCD, AO and BO are the bisectors of $\angle\text{A}$ and $\angle\text{B}$ respectively, $\angle\text{C} = 70^\circ$ and $\angle\text{D} = 30^\circ.$ Then, $\angle\text{AOB} =\ ?$
Answer
View full question & answer→- 50º
Solution:
It is given in the question that, ABCD is a quadrilateral where AO and BO are the bisectors of $\angle\text{A}$ and $\angle\text{B}.$
We know that, sum of all angles of a quadrilateral is equal to 360º
$∴ \angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$
$\angle\text{A} + \angle\text{B} + 70^\circ + 30^\circ = 360^\circ$
$\angle\text{A} + \angle\text{B} = 360^\circ - 100^\circ$
$\angle\text{A} + \angle\text{B} = 260^\circ$
$=\frac{1}{2}(\angle\text{A}+\angle\text{B})=\frac{1}{2}\times260^\circ$
$=\frac{1}{2}(\angle\text{A}+\angle\text{B})=160^\circ$
Now, in triangle AOB
$=\frac{1}{2}(\angle\text{A}+\angle\text{B})+\angle\text{AOB}=160^\circ$
$130^\circ + \angle\text{AOB} = 180^\circ$
$\angle\text{AOB} = 180^\circ - 130^\circ$
$\angle\text{AOB} = 50^\circ$
Question 1301 Mark
ABCD is a trapezium in which $\text{AB || DC}.$ M and N are the mid-points of AD and BC respectively. If AB = 12cm, MN = 14cm, then CD = ?
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Question 1311 Mark
If an angle of a parallelogram is two$-$third of its adjacent angle, the smallest angle of the parallelogram is:
Answer
View full question & answer→Let $x$ be one angle. Then $\text{x}+\frac{2}{3\text{x}}+23\text{x} =180$ um of the adjacent angles of a parallelogram is $180^\circ$
$x = 108,$ adjacent angle $= 180 - 108 = 72^\circ$
$x = 108,$ adjacent angle $= 180 - 108 = 72^\circ$
Question 1321 Mark
Write the correct answer in the following:
A diagonal of a rectangle is inclined to one side of the rectangle at 25º. The acute angle between the diagonals is:
A diagonal of a rectangle is inclined to one side of the rectangle at 25º. The acute angle between the diagonals is:
Question 1331 Mark
In a Quadrilateral ABCD, $\angle\text{A} = 90^\circ$ and AB = BC = CD = DA, Then ABCD is a:
Answer
View full question & answer→- Square
Solution:
A quadrilateral with pair of opposite sides equal and having one right angle is called Rectangle and a rectangle with all sides equal is called Square. So, ABCD is a Square.
Question 1341 Mark
Answer
View full question & answer→- x = 55º and y = 110º
Solution:
ABCD is a rectangle
The diagonals of a rectangle are congruent and bisect each other. Therefore, in $\triangle\text{AOB},$
we have:
OA = OB
$\angle\text{OAB} = \angle\text{OBA} = 35^\circ$
$\text{x} = 90^\circ – 35^\circ = 55^\circ$ and $\angle\text{AOB} = 180^\circ – (35^\circ + 35^\circ) = 110^\circ$
$\text{y} = \angle\text{AOB} = 110^\circ$ [Vertically opposite angles]
Hence, x = 55° and y = 110°
Question 1351 Mark
The diagonals AC and BD of a rectangle ABCD intersect each other at P. If $\angle\text{ABD}=50^\circ,$ then $\angle\text{DPC}=$
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Question 1361 Mark
If APB and CQD are two parallel lines, then the bisectors of $\angle\text{APQ},\angle\text{BPQ},\angle\text{CQP}$and $\angle\text{PQD}$ enclose a:
Answer
View full question & answer→- Rectangle.
Solution:
The bisectors of $\angle\text{APQ},\angle\text{BPQ},\angle\text{CQP}$ and $\angle\text{PQD}$ enclose a rectengle.
Question 1371 Mark
If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3 : 7 : 6 : 4, then ABCD is a:
Answer
View full question & answer→- Trapezium
Solution:
Let the angles be 3x, 7x, 6x, 4x
then 3x + 7x + 6x + 4x = 360
$\text{x}=\frac{360}{20}=18$
So angles are,
54º, 126º, 108º & 72º
Hence it is a trapezium.
Question 1381 Mark
The figure formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a square, only if:
Question 1391 Mark
Rhombus is a quadrilateral.
Answer
View full question & answer→- In which diagonals bisect opposite angles.
Solution:
Let ABCD be a rhombus.
Join BD which forms two triangles ABD and DCB. In $\triangle\text{ABD, AB = AD}.$
So, $\angle\text{ADB} = \angle\text{ABD}$ (angles opposite to equal sides are equal) ...(i)
But, $\angle\text{ABD} = \angle\text{BDC}$ and $\angle\text{ADB} = \angle\text{CBD}$ (alternate angles) ...(ii)
So, from (i) and (ii)
$\angle\text{ADB} = \angle\text{ABD}=\angle\text{BDC} = \angle\text{CBD}$
$\therefore$ diagonal BD bisects $\angle\text{B}$ and $\angle\text{D}.$
Question 1401 Mark
Answer
View full question & answer→- 40°
Solution:
$\text{AD || BC},$
$\Rightarrow\angle\text{DAO}=\angle\text{BCO}=30^{\circ}$ ...(Alternate angles)
$\Rightarrow\angle\text{BCO}=30^{\circ}$
$\angle\text{AOB}+\angle\text{BOC}=180^{\circ}$ ...(Linear pair of angles)
$\Rightarrow70+\angle\text{BOC}=180$
$\Rightarrow\angle\text{BOC}=110^{\circ}$
In $\triangle\text{CBO},$
$\angle\text{BOC}+\angle\text{BCO}+\angle\text{OBC}=180^{\circ}$ ...(Angle sum Property)
$\Rightarrow110+30+\angle\text{OBC}=180$
$\Rightarrow\angle\text{OBC}=40^{\circ}$
$\Rightarrow\angle\text{DBC}=40^{\circ}$ ...(D - O - B)
Question 1411 Mark
Which of the following quadrilateral is not a rhombus?
Answer
View full question & answer→- One angle between the diagonals is 60°
Solution:
For a rhombus, the angle between the diagonals is 90° and not 60°.
Question 1421 Mark
Answer
View full question & answer→- $\frac{1}{2}(\text{AB} - \text{CD})$
Solution:
Construction: Join CF and extent it to cut AB at point M
Firstly, in triangle MFB and triangle DFC
DF = FB (As F is the mid-point of DB)
$\angle\text{DFC} = \angle\text{MFB}$ (Vertically opposite angle)
$\angle\text{DFC} = \angle\text{FBM}$ (Alternate interior angle)
$∴$ By ASA congruence rule
$\triangle\text{MFB} ≅ \triangle\text{DFC}$
Now, in triangle CAM
E and F are the mid-points of AC and CM respectively.
$\therefore\ \text{EF}=\frac{1}{2}(\text{AM})$
$\text{EF}=\frac{1}{2}(\text{AB} - \text{MB})$
$\text{EF}=\frac{1}{2}(\text{AB} - \text{CD})$
Question 1431 Mark
In quadrilateral $\text{ABCD},$ if $\angle\text{A} = 60^\circ$ and $\angle\text{B} : \angle\text{C} : \angle\text{D} = 2 : 3 : 7,$ then $\angle\text{D}$ is:
Answer
View full question & answer→In quadrilateral, the sum of the all four angles equal to $360^\circ $. let $\angle\text{B} = 2\text{x},\ \angle\text{C} = 3\text{x}$ and $\angle\text{D} = 7\text{x}.$
$\angle\text{A} + \angle\text{B} + \angle\text{C} +\angle\text{D} = 360^\circ$
$60 + 2x + 3x + 7x = 360$
$12x = 300^\circ$
$x = 25^\circ$
So, $\angle\text{D} = 7\text{x} = 7(25^\circ) = 175^\circ$
$\angle\text{A} + \angle\text{B} + \angle\text{C} +\angle\text{D} = 360^\circ$
$60 + 2x + 3x + 7x = 360$
$12x = 300^\circ$
$x = 25^\circ$
So, $\angle\text{D} = 7\text{x} = 7(25^\circ) = 175^\circ$
Question 1441 Mark
In quadrilateral ABCD, if $\angle\text{A}= 60^\circ$ and $\angle\text{B}: \angle\text{C}: \angle\text{D} = 2:3:7,$ then $\angle\text{D}$ is:
Answer
View full question & answer→- 175º
Solution:
In quadrilateral, the sum of the all four angles equal to 360º.
Let $\angle\text{B} = 2\text{x}, \ \angle\text{C} = 3\text{x}$ and $\angle\text{D} = 7\text{x}.$
$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$
60 + 2x + 3x + 7x = 360
12x = 300
x = 25
So, $\angle\text{D} = 7\text{x} = 7 (25) = 175^\circ$
Question 1451 Mark
If area of a Parallelogram with sides ‘a’ and ‘b’ is A and that of a rectangle with sides ‘a’ and ‘b’ is B, then
Answer
View full question & answer→- A < B
Solution:
Area of Parallelogram = Base × Height
If 'a' is the side and 'b' is the base the height will be less than 'a' using Pythagoras theorem, a as Hypotenuse, h as height, A < B.
Question 1461 Mark
Answer
View full question & answer→- 2.8cm
Solution:
By using Mid-Point theorem,
DE = Half of BC
Hence, DE = 0.5 × 5.6 = 2.8cm
Question 1471 Mark
Diagonals of a quadrilateral ABCD bisect each other. If $\angle\text{A} = 45^\circ,$ then $\angle\text{B} =\ ?$
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Question 1481 Mark
The Diagonals AC and BD of a Parallelogram ABCD intersect each other at point O. If $\angle\text{DAC}=32^\circ$ and $\angle\text{AOB}=70^\circ,$ then $\angle\text{DBC}$ is equal to:
Answer
View full question & answer→- 38º
Solution:
$\angle\text{DAC} = \angle\text{ACB} = 32^\circ$ (alternate angles)
$\angle\text{AOB} + \angle\text{COB} = 180^\circ$ (linear pair)
$\angle\text{COB} = 180 - 70^\circ = 110^\circ$
In triangle BOC,
$\angle\text{BOC} + \angle\text{OCB} + \angle\text{CBO} = 180^\circ$ (angle sum property)
$110^\circ + 32^\circ+ \angle\text{CBO} = 180^\circ$
$\angle\text{CBO} = 180^\circ - 142^\circ = 38^\circ$
Question 1491 Mark
Length of the line segment joining the mid-points of two sides of a triangle is ________ the third side of the Triangle.
Answer
View full question & answer→- Half the length of.
Solution:
According to mid point theorem, A line segment joining mid points of any two sides of a triangle is parallel to third side and length of that line segment is half of the third side.
Question 1501 Mark
In a trapezium$ \text{ABCD},$ if $AB || CD,$ then $(AC^2 + BD^2) =$ ?
Answer
Construction: Draw perpendicular from $D$ and $C$ on $AB$ which meets $AB$ at $E$ and $F,$ respectively.
So, $\text{DEFC}$ is a parallelogram, since one pair of opposite sides are parallel and equal.
In $\triangle\text{ABC},$
$\angle\text{B}$ is an acute angle.
$\Rightarrow\text{AC}^2=\text{BC}^2+\text{AB}^2-2\text{AB}\times\text{AE}$
In $\triangle\text{ABD},$
$\angle\text{A}$ is an acute angle.
$\Rightarrow\text{BD}^2=\text{AD}^2+\text{AB}^2-2\text{AB}\times\text{AF}$
$\Rightarrow\text{AC}^2+\text{BD}^2=\text{BC}^2+\text{AD}^2+2\text{AB}(\text{AB}-\text{BE}-\text{AF})$
$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{EF}$
$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{CD}$
View full question & answer→Construction: Draw perpendicular from $D$ and $C$ on $AB$ which meets $AB$ at $E$ and $F,$ respectively.
So, $\text{DEFC}$ is a parallelogram, since one pair of opposite sides are parallel and equal.
In $\triangle\text{ABC},$
$\angle\text{B}$ is an acute angle.
$\Rightarrow\text{AC}^2=\text{BC}^2+\text{AB}^2-2\text{AB}\times\text{AE}$
In $\triangle\text{ABD},$
$\angle\text{A}$ is an acute angle.
$\Rightarrow\text{BD}^2=\text{AD}^2+\text{AB}^2-2\text{AB}\times\text{AF}$
$\Rightarrow\text{AC}^2+\text{BD}^2=\text{BC}^2+\text{AD}^2+2\text{AB}(\text{AB}-\text{BE}-\text{AF})$
$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{EF}$
$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{CD}$