Question

in the given fig. abcd is a rectangle p is midpoint of cd . if qb =7cm and ad=9cm and dc=24cm . then prove angle aqp is 90degrees​

Answer 1

Answer:

Step-by-step explanation:

Given 1hal ABCO 1s a rec1ang1e.

Now.DC • AB • 24 cm (Opposesldesolte<:!Mgleare equal)

Now. each engleof rectangleISangNangle.

So, LA = LB • LC = LD = 90•

Now, LOSA + LABC • 180° (Linear pa11)

•LOSA • 90'

In LIOBA

Acr' = Ae2 + OB' (Pylrlago<as lhe«em)

-Acr s 24' + 71 • 576 + 49 = 625

•AO = 25 cm

Now, AO • BC • 9 cm

Nmv, P rs the mid pornIof DC, then

OP = PC • 12cm

ln llOCP

Pa2 = oc> + PC' (Pythagoras theorem)

=162 + 122

= 256+ 144

= 400

-PO = 20an

In LIAOP,

AP' = Ao' + OP'

=92 + 122

=81 + 144

=225

So,AP = 15cm In LiAPQ,

AQ2 = 252 = 625

AP' + P02 = 152 + 202 = 225 + 400 = 625

Since, AP' + P02 = Acl-

So,by converse of Py1hagoras theorem, LAPO = 90°