Mensuration
- If O is the centroid and AD, BE and CF are the three medians of ∆ABC with an area of 96 cm² then the area of ∆BOD in cm² is
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Point O is the centroid of ∆ABC.
∴ ∆AOB ≡ ∆AOC ≡ ∆BOC
Again, ∆BOD ≡ ∆COD∴ Area of ∆BOD = 1 × Area of ∆ABC 6 = 1 × 96 = 16 sq.cm. 6 Correct Option: C
Point O is the centroid of ∆ABC.
∴ ∆AOB ≡ ∆AOC ≡ ∆BOC
Again, ∆BOD ≡ ∆COD∴ Area of ∆BOD = 1 × Area of ∆ABC 6 = 1 × 96 = 16 sq.cm. 6
- In a triangle ABC, AB = 8 cm, AC = 10 cm and ÐB = 90°, then the area of ∆ABC is
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AC = 10 cm.
AB = 8 cm.
∴ BC = √AC² - AB²
= √10² - 8² = √100 - 64
= √36 = 6 cm.∴ Area of ∆ABC = 1 × AB × BC 2 ∴ Area of ∆ABC = 1 × 8 × 6 = 24 sq. cm. 2 Correct Option: D
AC = 10 cm.
AB = 8 cm.
∴ BC = √AC² - AB²
= √10² - 8² = √100 - 64
= √36 = 6 cm.∴ Area of ∆ABC = 1 × AB × BC 2 ∴ Area of ∆ABC = 1 × 8 × 6 = 24 sq. cm. 2
- In figure, DE || BC. If DE = 3 cm, BC = 6 cm and area of ∆ADE = 15 sq. cm, then the area of ∆ABC is
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DE || BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
By AA–similarity,
∆ADE ~ ∆ABC∴ Area of ∆ABC = BC² Area of ∆ADE DE² ⇒ Area of ∆ABC = 6² 15 3² = 36 = 4 9
∴ Area of DABC = 4 × 15
= 60 sq. cm.Correct Option: D
DE || BC
∴ ∠ADE = ∠ABC
∠AED = ∠ACB
By AA–similarity,
∆ADE ~ ∆ABC∴ Area of ∆ABC = BC² Area of ∆ADE DE² ⇒ Area of ∆ABC = 6² 15 3² = 36 = 4 9
∴ Area of DABC = 4 × 15
= 60 sq. cm.
- ∆ABC is a right angled triangle, the radius of its circumcircle is 3 cm and the length of its altitude drawn from the opposite vertex to the hypotenuse is 2 cm. Then the area of the triangle is
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The angle in a semi–circle is a right angle.
∴ BC = 2 × 3 = 6 cm.
OA = 2 cm.∴ Area of ∆ABC = 1 × BC × OA 2 = 1 × 6 × 2 = 6 sq.cm. 2 Correct Option: C
The angle in a semi–circle is a right angle.
∴ BC = 2 × 3 = 6 cm.
OA = 2 cm.∴ Area of ∆ABC = 1 × BC × OA 2 = 1 × 6 × 2 = 6 sq.cm. 2
- The lengths of the diagonals of a rhombus are 8 cm and 6 cm. The area of rhombus is :
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Area of the rhombus
∴ Area of ∆ABC = 1 × d1 × d2 2 = 1 × 8 × 6 sq. cm. = 24 sq.cm. 2 Correct Option: D
Area of the rhombus
∴ Area of ∆ABC = 1 × d1 × d2 2 = 1 × 8 × 6 sq. cm. = 24 sq.cm. 2