You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Mathematics

Equal Angles Subtend Equal Sides

Consider a sq...

Question

Consider a square $A B C D$ of side $12$ and let $M, N$ be the midpoints of $A B, C D$ respectively. Take a point $P$ on MN and let $A P = r, P C = s$ . Then the area of the triangle whose sides are $r, s, 12$ is

72

No worries! We‘ve got your back. Try BYJU‘S free classes today!

36

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{r s}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{r s}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $36$
Since PD=PC=s,PA=r,AD=12
Therefore we will find the area of $Δ$ APD
$Δ A P D = \frac{1}{2} \times 12 \times 6$
$= 36$

Suggest Corrections

Similar questions

Q. Let

A B C D

be a square of a side length

1

. Let

P, Q, R, S

be points in the interiors of the sides

A D, B C, A B, C D

respectively, such that

P Q

and

R S

intersect at right angles. If

P Q = \frac{3 \sqrt{3}}{4}

then

R S

equals

Let ABCD be a square of a side length l, Let P, Q, R, S be points in the interiors of the sides AD, BC, AB, CD, respectively, such that PQ and RS intersect at right angles. If PQ = $\frac{3 \sqrt{3}}{4}$ , then RS equals

Q. Let

A B C D

be a square of side length

1.

Let

P, Q, R, S

be points in the interiors of the sides

A D, B C, A B, C D,

respectively, such that

P Q

and

R S

intersect at right angles. If

P Q = \frac{3 \sqrt{3}}{4}

then

R S

equals

Let ABCD be a square. Locate points P, Q, R, S on the sides AB, BC, CD, DA respectively such that AP = BQ = CR = DS. Prove that PQRS is a square.

Join BYJU'S Learning Program

Consider a square ABCD of side 12 and let M,N be the midpoints of AB,CD respectively. Take a point P on MN and let AP=r,PC=s. Then the area of the triangle whose sides are r,s,12 is

The correct option is B 36Since PD=PC=s,PA=r,AD=12Therefore we will find the area of ΔAPDΔAPD=12×12×6=36

Consider a square $A B C D$ of side $12$ and let $M, N$ be the midpoints of $A B, C D$ respectively. Take a point $P$ on MN and let $A P = r, P C = s$ . Then the area of the triangle whose sides are $r, s, 12$ is

The correct option is B $36$
Since PD=PC=s,PA=r,AD=12
Therefore we will find the area of $Δ$ APD
$Δ A P D = \frac{1}{2} \times 12 \times 6$
$= 36$