Question

ΔABC is an isosceles right angled triangle, in which a rectangle is inscribed in such a way that the length of the rectangle (RQ) is twice of breadth (PQ). Q and R lie on the hypotenuse and P, S lie on the two different smaller sides of the triangle. What is the ratio of the areas of the rectangle and that of triangle.

Answer 1

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RC

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]In ΔARQ and ΔQPB

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]In ΔARQ and ΔQPBAR=BP

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]In ΔARQ and ΔQPBAR=BP∠ARQ=∠QPB=90

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]In ΔARQ and ΔQPBAR=BP∠ARQ=∠QPB=90 o

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]In ΔARQ and ΔQPBAR=BP∠ARQ=∠QPB=90 o

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]In ΔARQ and ΔQPBAR=BP∠ARQ=∠QPB=90 o QR=PQ

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]In ΔARQ and ΔQPBAR=BP∠ARQ=∠QPB=90 o QR=PQ∴ΔARQ≅ΔQPB

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]In ΔARQ and ΔQPBAR=BP∠ARQ=∠QPB=90 o QR=PQ∴ΔARQ≅ΔQPB⇒AQ=QB

Given: ΔABC is an isosceles right triangle and square CPQR is inscribed in it.CPQR is a square∴CP=PQ=PR=RCΔABC is an isosceles triangle∴AC=BC⇒AR+RC=CP+BP⇒AR=BP ……..(1) [∵RC=CP]In ΔARQ and ΔQPBAR=BP∠ARQ=∠QPB=90 o QR=PQ∴ΔARQ≅ΔQPB⇒AQ=QB∴ Q bisects the hypotenuses AB.

Answer 2

$\large \red {\mathbb{Answer:-}}$

$\large \bf \blue{given}$

ΔABC is an isosceles right triangle and square CPQR is inscribed in it.

CPQR is a square

∴CP=PQ=PR=R

ΔABC is an isosceles triangle

∴AC=BC

⇒AR+RC=CP+BP

⇒AR=BP ……..(1) [∵RC=CP]

In ΔARQ and ΔQPB

AR=BP

∠ARQ=∠QPB=90

o

QR=PQ

∴ΔARQ≅ΔQPB

⇒AQ=QB

∴ Q bisects the hypotenuses AB.

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