# In Fig. 10.23, PQRS is a square and SRT is an equilateral triangle. Prove that (i) PT = QT (ii) ∠TQR = 15°

It is given that

Δis a square and Δ is an equilateral triangle.

We have to prove that

(1) and (2)

(1)

Since,

(Angle of square)

(Angle of equilateral triangle)

Similarly, we have

Thus in and we have

(Side of square)

And (equilateral triangle side)

So by congruence criterion we have

Hence.

(2)
Since
QR = RS ( Sides of Square)
RS = RT (Sides of Equilateral triangle)

We get
QR = RT

Thus, we get
$\angle TQR=\angle RTQ$  (Angles opposite to equal sides are equal)

Now, in the triangle TQR, we have

$\angle TQR+\angle RTQ+\angle QRT={180}^{0}\phantom{\rule{0ex}{0ex}}\angle TQR+\angle TQR+{150}^{0}={180}^{0}\phantom{\rule{0ex}{0ex}}2\angle TQR+{150}^{0}={180}^{0}\phantom{\rule{0ex}{0ex}}2\angle TQR={180}^{0}-{150}^{0}\phantom{\rule{0ex}{0ex}}2\angle TQR={30}^{0}\phantom{\rule{0ex}{0ex}}\angle TQR=\frac{{30}^{0}}{2}={15}^{0}\phantom{\rule{0ex}{0ex}}$

• -1
What are you looking for?