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ABCD is a parallelogram with sides AB = 12 cm, BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram. Also find the distance between its shorter sides.

Mensuration

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Answer

In triangle ABC,

ABCD is a parallelogram with sides AB = 12 cm, BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram. Also find the distance between its shorter sides. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Let,

BC = a = 10 cm, AC = b = 16 cm and AB = c = 12 cm.

We know that,

Semi-perimeter (s) = a+b+c2\dfrac{a + b + c}{2}

= 10+16+122=382\dfrac{10 + 16 + 12}{2} = \dfrac{38}{2} = 19 cm.

By Heron's formula,

Area of triangle =s(sa)(sb)(sc)=19(1910)(1916)(1912)=19×9×3×7=3591=59.9 cm2.\text{Area of triangle } = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{19(19 - 10)(19 - 16)(19 - 12)} \\[1em] = \sqrt{19 \times 9 \times 3 \times 7} \\[1em] = \sqrt{3591} \\[1em] = 59.9 \text{ cm}^2.

We know that,

Diagonal of a parallelogram divides it into two triangles of equal area.

∴ Area of triangle ABC = Area of triangle ADC

∴ Area of parallelogram = 2 × Area of triangle ABC.

= 2 × 59.9

= 119.8 cm2.

Let DM be the distance between the shorter sides of the parallelogram.

By formula,

Area of parallelogram = base × height = BC × DM

Substituting the values we get,

⇒ 119.8 = 10 × DM

⇒ DM = 119.810\dfrac{119.8}{10}

⇒ DM = 11.98 cm.

Hence, the distance between shorter sides = 11.98 cm and area of parallelogram = 119.8 cm2.

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