Find the perimeter and area of quadrilateral ABCD in which AB = 9 cm, AD = 12 cm, BD = 15 cm, CD = 17 cm and ∠CBD = 90°.

Given,

AB = 9 cm

AD = 12 cm

BD = 15 cm

CD = 17 cm

∠CBD = 90°

Diagonal BD divides the quadrilateral into two triangles: △ABD and △BCD.

For △ABD,

Sides are 9 cm, 12 cm, 15 cm.

Since, 9² + 12² = 81 + 144 = 225 = 15².

So they are the pythagorean triplets.

So triangle △ABD is a right angled triangle.

Area of right angle triangle = $\dfrac{1}{2}$ × Base × Height

Area of triangle ABD = $\dfrac{1}{2}$ × AB × AD

= $\dfrac{1}{2}$ × 9 × 12

= 9 × 6

= 54 cm².

For △BCD,

Since, ∠CBD = 90°

∴ Applying Pythagoras theorem for the △BCD,

⇒ BD² + BC² = DC²

⇒ 15² + BC² = 17²

⇒ BC² = 289 - 225

⇒ BC² = 64

⇒ BC = $\sqrt{64}$

⇒ BC = 8 cm.

Area of △BCD = $\dfrac{1}{2}$ × BC × BD

= $\dfrac{1}{2}$ × 8 × 15

= 4 × 15

= 60 cm².

Area of quadrilateral △ABCD = Area of △ABD + Area of △BCD

= 54 + 60

= 114 cm².

Perimeter of quadrilateral ABCD = AB + BC + CD + DA

= 9 + 8 + 17 + 12

= 46 cm.

Hence, area = 114 cm² and perimeter = 46 cm.