Question
The given figure shows the parallelograms $\text{ABCD}$ and $\text{APQR}$.Show that these parallelograms are equal in the area.$[$ Join $B$ and $R ]$
✓
Answer
Join $B$ and $R$ and $P$ and $R.$
We know that the area of the parallelogram is equal to twice the area of the triangle if the triangle and the parallelogram are on the same base and between the parallels.
Consider $\text{ABCD}$ parallelogram:Since the parallelogram $\text{ABCD}$ and the $\triangle ABR$ lie on $AB$ and between the parallels $AB$ and $DC$, we have
Area$(\square ABCD)=2 \times$Area$(\triangle ABR) \dots....(1)$
We know that the area of triangles with the same base and between the same parallel lines are equal.
Since the triangles $\text{ABR}$ and $\text{APR}$ lie on the same base $AR$ and between the parallels $AR$ and $QP,$ we have,
Area $(\triangle ABR) =$ Area $(\triangle APR) \dots....(2)$
From equations $(1)$ and $(2)$, we have,
Area$(\square ABCD )=2 \times$ Area$(\triangle APR )\dots .....(3)$
Also, the $\triangle APR$ and the parallelograms, $AR$ and $QR,$ lie on the same base $AR$ and between the parallels, $AR$ and $QP,$
Area$(\triangle APR)=\frac{1}{2} \times$ Area$(\square ARQP)\dots....(4)$
Using $(4)$ in equation $(3),$ We have,
Area$(\square \text{ABCD})=2 \times \frac{1}{2} \times$ Area$(\square \text{ARQP})$
Are$a(\square \text{ABCD})=$Area$(\square \text{ARQP})$
Hence Proved.
We know that the area of the parallelogram is equal to twice the area of the triangle if the triangle and the parallelogram are on the same base and between the parallels.
Consider $\text{ABCD}$ parallelogram:Since the parallelogram $\text{ABCD}$ and the $\triangle ABR$ lie on $AB$ and between the parallels $AB$ and $DC$, we have
Area$(\square ABCD)=2 \times$Area$(\triangle ABR) \dots....(1)$
We know that the area of triangles with the same base and between the same parallel lines are equal.
Since the triangles $\text{ABR}$ and $\text{APR}$ lie on the same base $AR$ and between the parallels $AR$ and $QP,$ we have,
Area $(\triangle ABR) =$ Area $(\triangle APR) \dots....(2)$
From equations $(1)$ and $(2)$, we have,
Area$(\square ABCD )=2 \times$ Area$(\triangle APR )\dots .....(3)$
Also, the $\triangle APR$ and the parallelograms, $AR$ and $QR,$ lie on the same base $AR$ and between the parallels, $AR$ and $QP,$
Area$(\triangle APR)=\frac{1}{2} \times$ Area$(\square ARQP)\dots....(4)$
Using $(4)$ in equation $(3),$ We have,
Area$(\square \text{ABCD})=2 \times \frac{1}{2} \times$ Area$(\square \text{ARQP})$
Are$a(\square \text{ABCD})=$Area$(\square \text{ARQP})$
Hence Proved.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find $x$ :
→ A trader fixes the selling price of his goods at $50\%$ above the cost price. He sells half of his stock at this price, a quarter of his stock at a discount of $20\%$ on the original selling price, and the rest at a discount of $36\%$ on the original selling price. Find the gain per cent altogether.
→A manufacturer makes $800$ straws at a cost of $50$ paise per straw. He fixes the selling price such that if only $640$ straws are sold, he would make a profit of $50\%$ on his outlay. Find his actual profit or loss percent, if he sells $720$ straws.
→From the following figure, find the values of:$(i) \sin B;(ii) \tan C;(iii)\sec^2 B - \tan^2B;(iv)\sin^2C + \cos^2C$
→Plot point $A(5, -7)$. From point $A$, draw $AM$ perpendicular to the $x-$axis and $AN$ perpendicular to the $y-$axis. Write the coordinates of points $M$ and $N.$
→In $\triangle ABC, MN$ is drawn parallel to $BC.$ If $AB = 3.5\ cm, AM : AB = 5 : 7$ and $NC = 2\ cm,$ find:$(i) AM;(ii) AC$
→Two oils $A$ and $B$ are mixed in the ratio $3: 2$. The cost price of oil $A$ and $B$ are $Rs. 300$ per litre and $Rs.400$ per litre respectively. If one-fourth of the mixture is sold at $Rs. 450$ per litre and the remaining of the mixture at $Rs. 500$ per litre, find the profit per cent on the whole.
→In the given figure, the lengths of arcs $AB$ and $BC$ are in the ratio $3:2.$ If $\angle AOB = 96^\circ $,find: $(i) \angle BOC ;(ii) \angle ABC$
→The cost of fencing a triangular field at the rate of $Rs.15$ per m is $Rs.900$. If the lengths of the triangle are in the ratio $3:4:5$, find the area of the triangle and the cost of cultivating it at $Rs.48$ per $kg\ sq.m.$
→The sum of the digits of the digits of two digit number is $5$. If the digits are reversed, the number is reduced by $27$. Find the number.
→