3 Marks Question

Question 13 Marks

Construct a triangle with sides $5 cm, 5.5 cm$ and 6.5 cm . Now construct another triangle, whose sides are $\frac{3}{5}$ times the corresponding sides of the given triangle.

Answer

Step 1: Construct a triangle $P Q R$ such that $Q R=5.5 cm$. Considering $Q$ as center, mark an arc of length 5 cm and considering $R$ as center, mark an arc of length 6.5 cm . The point of intersection of ares is called $P$.
Step 2: Draw a line $Q M$ such that it makes an acute angle with $Q R$.
Step 3: Draw 5 points $Q_1, Q_2, Q_3, Q_4$ and $Q_5$ on equal distance from each other on $Q M$ and then join the fifth mark to point $R$.
Step 4: Draw a line $Q_3 R$ ' parallel to $Q_5 R$ from $Q_3$

Step 5: Draw a line $R$ ' $P$ ' parallel to $R P$

Step 6: $P ^{\prime} Q R$ ' is the required triangle.$A ^{\prime} BC ^{\prime}$ is the required triangle.

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Question 23 Marks

Draw a right triangle ABC in which $A B=6 cm$, $B C=8 cm$ and $\angle B=90^{\circ}$. Draw $B D$ perpendicular from $B$ on $A C$ and draw a circle passing through the points $B, C$ and $D$. Construct tangents from $A$ to this circle.

Answer

Follow the given steps to construct the figure. Step 1:Draw a line $A B=6 cm$ segment from point B , draw a ray making an angle of $90^{\circ}$ with AB . Now with $B$ as center and radius 8 cm draw an arc cutting the ray at point C. Join AC, to form $\triangle A B C$. Thus, $\triangle A B C$ is created.
Step 2: Bisect BC and name the midpoint of BC as E . So , the center of circle is E .
Step 3: Join points A and E. Bisect AE and name the midpoint of AE is M .
Step 4: With M as centre and ME as radius, draw a circle.
Step 5: Let it intersect given circle at $B$ and $P$.
Step 6: Join AP and AB.
Here, AB and AP are the required tangents to the circle from A

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Question 33 Marks

Construct a triangle with sides $5 cm, 4 cm$ and 6 cm . Then construct another triangle whose sides are $\frac{2}{3}$ times the corresponding sides of first triangle.

Answer

Following are the steps of constructing a required triangle.
Step 1: Draw a line segment $A B=4 cm$ and then draw an are of radius 5 cm considering $A$ as a center.
Step 2: Draw an arc of radius 6 cm considering $B$ as center.
Step 3: Name the point where both the ares from step 1 and step 2 intersect as $C$.
Step 4: Join $B C$ and $A C . \triangle A B C$ is formed.
Step 5: Draw a ray $A Y$ which forms an acute angle with $A B$ on the opposite side of the vertex $C[1 / 2]$ Step 6: Locate three points $A_1, A_2$, and $A_3$ on lines $A Y$ such that $A A 1=A_1 A_2=A_2 A_3$.
Step 7: Join $B$ and $A_3$.
Step 8: Draw a line segment from point $A_2$ and parallel to $B A_3$ which intersects $A B$ at point $B$ '.
Step 9: Draw a line segment from $B^{\prime}$ and parallel to $B C$ which intersects $A C$ at the point $C^{\prime}$. [1/2] Thus, required triangle $\triangle A B^{\prime} C^{\prime}$ whose sides are $\frac{2}{3}$ times the corresponding sides of first triangle ( $\triangle A B C$ ) has been constructed.

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Question 43 Marks

Draw a triangle $A B C$ with $B C=7 cm, \angle B =45^{\circ}$ and $\angle C =60^{\circ}$. Then construct another triangle, whose sides are $\frac{3}{5}$ times the corresponding sides of $\triangle A B C$.

Answer

In order to construct the given triangle, follow the following steps:
Step 1: Draw $B C=7 cm$.
Step 2: At $B$, construct $\angle B=45^{\circ}$ and at $C$, construct $\angle C=60^{\circ}$. They intersect each other at $A$. Thus, $\triangle A B C$ is constructed.
Step 3: Construct an acute angle $\angle C B Z$ at $B$ on opposite side of vertex $A$ of $\triangle A B C$
Step 4: Along $B Z$, mark off 5 points $B_1, B_2, B_3, B_4$, $B_5$ such that
$
B B_1=B_1 B_2=B_2 B_3=B_3 B_4=B_4 B_5 .
$
Step 5: Join $B_5 C$
Step 6: Since we have to construct a triangle each of whose sides is $\frac{3}{5}$ of the corresponding sides of $\triangle A B C$. So, take three parts out of five equal parts on $B Z$ i.e., from $B_3$, draw $B_3 C| | B_5 C$, meeting $B C B C$ at $C$.
Step 7: From $C^{\prime}$, draw $C^{\prime} A^{\prime} \mid I C A$, meeting $B A$ at $A^{\prime}$. Thus, $A^{\prime} B C^{\prime}$ is the required triangle, each of whose sides are $\frac{3}{5}$ of the corresponding sides of $\triangle A B C$ as shown below

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Question 53 Marks

Draw a triangle $A B C$ in which $AB =5 cm$, $B C=6 cm$ and $\angle A B C=60^{\circ}$. Then construct a triangle whose sides are $\frac{5}{7}$ times the corresponding sides of $\triangle A B C$.

Answer

1. Steps of construction for triangle $A B C$ :
(1) Draw a line segment $B C=6 cm$
(2) From point $B$, draw a ray making an angle of $60^{\circ}$ with $B C$.
(3) Now with $B$ as center and radius 5 cm draw an are cutting the ray at point $A$.
(4) Join $A C$, to form $\triangle A B C$.
Now steps of construction for similar triangle,
(1) Draw a ray $B X$ making an acute angle opposite to vertex $A$.
(2) Mark 7 points $B_1, B_2, B_3, B_4, B_5, B_6, B_7$ at equal distance from each other on $B X$ such that
$
B B_1=B_1 B_2=B_2 B_3=B_3 B_4=B_4 B_5=B_5 B_6=B_6 B_7
$
(3) Join the point $C$ and $B_7$ and draw $B_5 C$ 'parallel to $B_7 C$.
(4) Draw $C$ 'A' parallel to $C A$.
Hence $A^{\prime} B^{\prime} C$ is the required triangle as shown below.

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Question 63 Marks

Construct a pair of tangents to a circle of radius 4 cm from a point $P$ lying outside the circle at a distance of 6 cm from the centre.

Answer

Steps of Construction:
1. Construct a circle of radius 4 cm .

2. Take a point $P, 6 cm$ away from the centre of the circle.
3. Join $O P$.
4. Draw a perpendicular bisector of $O M$ which intersects $O P$ at $M$.
5. Taking $M$ as a centre and $O M$ as radius, draw another circle which intersects the previous circle at $A$ and $B$.
6. Join $P A$ and $P B$.
$P A$ and $P B$ are the required tangents.

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Question 73 Marks

Draw a line segment of length 7 cm and divide it in the ratio $5: 3$.

Answer

Steps of construction:
1. Draw a line segment $A B$ of length 7 cm .
2. Draw a ray AX making an acute angle with AB .
3. Mark 8 (i.e., $5+3$ ) points as $A _1, A_2, A_3 \ldots A_8$ on AX such that $AA _1= AA _2= AA _3 \ldots= AA _7= AA _8$
4. Join $BA _8$
5. Through $A _5$ (since we need 5 parts to 8 parts) draw $CA _5$ parallel to $BA _8$, where C lies on AB .
Now, $AC : CB =5: 3$

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Question 83 Marks

Construct a triangle with its sides $4 cm, 5 cm$ and 6 cm . Then construct a triangle similar to it whose sides are $\frac{2}{3}$ of the corresponding sides of the first triangle.

Answer

In $\triangle ABC$, $
\begin{aligned}
AB & =4 cm \\
BC & =6 cm \\
AC & =5 cm \\
\text { Ratio } & =\frac{2}{3}
\end{aligned}
$

$A ^{\prime} BC ^{\prime}$ is the required triangle.

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Question 93 Marks

Draw a circle of radius 2.5 cm . Take a point $P$ at a distance of 8 cm from its centre. Construct a pair of tangents from the point $P$ to the circle.

Answer

PA and PB are rquired tangents

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Maths 3 Marks Question

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