Question

please answer this question then i will mark as brainlist​

Answer 1

Answer:

1)

GIVEN-

ABCD is a rectangle

AB=2cm

BC=6cm

TO FIND-

Ar. of rec ABCD

Measure of angle B

Ar. of triangle ABC

TO PROVE-

Sides of triangle ABD are equal to sides pf triangle ACD.

SOLUTION-

Ar. of rec. = l×b

Ar. of rec. ABCD=

$2 \times 6 {cm}^{2} \\ = > 12 {cm}^{2}$

Measure of angle B=90 degrees [As ABCD is a rec. and in a rectangle all angles are 90 degrees]

Ar. of triangle=

$\frac{1}{2} \times b \times h$

Therefore Ar. of triangle ABC=

$\frac{1}{2} \times 2 \times 6 \: {cm}^{2} \\ \\ = > 6 {cm}^{2}$

PROOF-

AB=CD [ABCD is a rectangle]

BC=AD [ABCD is a rectangle]

AC=AC [Common]

Therefore, Triangle ABC is congruent to triangle ACD

Therefore, by CPCT sides os triangle ABC = sides of triangle ACD

2)

GIVEN-

PQRS is a rectangle

PS=3cm

PQ=10cm

TO FIND-

Ar. of rec PQRS

Ar. of triangle QPS

Ar. of triangle PTQ

Ar. of triangle PUQ

SOLUTION-

Ar. of rec. = l×b

Ar. of rec. PQRS=

$3 \times 10 {cm}^{2} \\ = > 30 {cm}^{2}$

Ar. of triangle=

$\frac{1}{2} \times \: b \times h$

Therefore Ar. of triangle QPS [In this case height will be PS as triangle QPS is a right angled triangle because PQRS is a rectangle] =

$\frac{1}{2} \times 10 \times 3 {cm}^{2} \\ \\ = > 15 {cm}^{2}$

Ar. of triangle PTQ [ the explanation of how the heightis 3 cm is in the pic above]=

$\frac{1}{2} \times 10 \times 3 {cm}^{2} \\ \\ = > 15 {cm}^{2}$

Ar. of triangle PUQ[the explanation of how the heightis 3 cm is in the pic above]=

$\frac{1}{2} \times 10 \times 3 {cm}^{2} \\ \\ = > 15 {cm}^{2}$