In trapezium A B C D , A B ∥ D C , D C = 2 A B . E F drawn parallel to A B cuts A D in F and B C in E such that B E E C = 3 4 . Diagonal D B intersects E F at G . Prove that 7 F E = 10 A B .

Given : A trapezium A B C D with A B ∥ D C and D C = 2 A B . Given that E F drawn parallel to A B cuts A D in F and B C in E such that B E E C = 3 4 . Diagonal D B intersects E F at G . Consider ∆ B D C and ∆ B G E : ∠ 1 = ∠ 2 (Corresponding angles) ∠ 3 = ∠ 3 (Common Angle) ∆ B D C ~ ∆ B G E (By AA Criterion Similarity) B E B C = G E D C (Corresponding sides of similar triangles are proportional) So, G E D C = 3 7 G E 2 A B = 3 7 G E A B = 6 7 In ∆ B D C , G E ∥ D C Using Basic Proportionality theorem, B E E C = B G G D ......(1) In ∆ A D B , G F ∥ A B Using Basic Proportionality theorem, A F F D = B G G D ......(2) From (1) and (2), we get B E E C = A F F D = 3 4 In ∆ D G F and ∆ D B A , ∠ 4 = ∠ 5 (Corresponding angles) ∠ 6 = ∠ 6 (Common) ∆ D G F ~ ∆ D B A (By AA Criterion Similarity) Then G F A B = D F A D (Corresponding sides of similar triangles are proportional) G F A B = 4 7 E F A B = G E A B + G F A B = 6 7 + 4 7 = 10 7 Thus, 7 F E = 10 A B Hence, proved.

If a sinθ + b cosθ = c , then prove that a cosθ - b sinθ = ± a 2 + b 2 - c 2 .

Given: a sinθ + b cosθ = c Squaring on both sides, we get; a sinθ + b cosθ 2 = c 2 Using the algebaric identity a + b 2 = a 2 + 2 a b + b 2 , we get: a 2 sin 2 θ + b 2 cos 2 θ + 2 a b sinθcosθ = c 2 Using the identity sin 2 θ + cos 2 θ = 1 , we get: a 2 1 - cos 2 θ + b 2 1 - sin 2 θ + 2 a b sinθcosθ = c 2 a 2 - a 2 cos 2 θ + b 2 - b 2 sin 2 θ + 2 a b sinθcosθ = c 2 a 2 + b 2 - c 2 = a 2 cos 2 θ + b 2 sin 2 θ - 2 a b sinθcosθ Using the identity a - b 2 = a 2 - 2 a b + b 2 , we get: a 2 + b 2 - c 2 = a cosθ - b sinθ 2 a cosθ - b sinθ = ± a 2 + b 2 - c 2 Hence, proved.

Prove that tanθ 1 - tanθ - cotθ 1 - cotθ = cosθ + sin θ cosθ - sin θ .

We need to prove : tanθ 1 - tanθ - cotθ 1 - cotθ = cosθ + sin θ cosθ - sin θ We know that, tanθ = sinθ cosθ and cotθ = cosθ sinθ . sinθ cosθ 1 - sinθ cosθ - cosθ sinθ 1 - cosθ sinθ sinθ cosθ cosθ - sinθ cosθ - cosθ sinθ sinθ - cos θ sinθ sinθ cosθ - sinθ - cosθ sin θ - cos θ cosθ + sin θ cosθ - sin θ Hence, proved.

E M B I B E

New Mathematics for Class X>Chapter 16 - Model Test Paper>SECTION>Q 23

MEDIUM

10th CBSE

IMPORTANT

Earn 100

The angry Arjun carried some arrows for fighting with Bheeshm. With half the arrows, he cut down the arrows thrown by Bheeshm on him. Six arrows, he used to kill the rath driver of Bheeshm. With one arrow each he knocked down respectively the rath, flag and the bow of Bheeshm. Finally, with one more than four times the square root of arrows he laid Bheeshm unconscious on an arrow bed. Find the total number of arrows Arjun had.

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Important Questions on Model Test Paper

MEDIUM

10th CBSE

IMPORTANT

New Mathematics for Class X>Chapter 16 - Model Test Paper>SECTION>Q 23

A peacock is sitting on top of a pillar, which is

9 m

high. From a point

27 m

away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing this, the peacock pounces on the snake. If their speeds are equal, at what distance from hole is the snake caught?