Question

in the given figure triangle ABC similar to triangle PQR find the value of Y+Z

Answer 1

since both triangles are similar
so by BPT
pq/ab = pr/bc= qr/ac
so first we can take
pq/ab=pr/bc
3/z = 6/ 8
z= 4cm

pq/ab = qr/ac
3/4 = y/4✓3
y=3✓3

z+y = 4cm + 3✓3cm

Answer 2

Given:

• ΔPQR ≅ ΔABC

To Find:

• The values of x and y.

Solution:

• From the figure, we get to know that ΔPQR is a right-angled triangle.
• Using the Pythagoras theorem, $PR^2 = QR^2+PQ^2$
• ⇒ $(6)^2 = y^2+(3)^2$
• Substituting the values,
• ⇒ 36 = $y^2$ + 9
• $y^2 = 36-9$  ⇒ y = $\sqrt{27}$ = $\sqrt{9*3}$  = $3\sqrt{3}$
• y = $3\sqrt{3}$  cm
• Consider the second ΔABC which is a right-angle triangle,
• Again using the Pythagoras, $BC^2 = AB^2+AC^2$
• ⇒ $(8)^2 = z^2+(4\sqrt{3})^2$
• Substituting the values,
• ⇒ $64 = z^2 + 48$
• ⇒ $z^2 = 64-48$
• z = $\sqrt{64-48}$ = $z = \sqrt{16}$ = 4
• z = 4 cm

The Values of y and z are $3\sqrt{3}$ cm and 4 cm respectively.