Sides AB and BE of a right triangle, right angled at B are of lengths 16 cm and 8 cm respectively. The length of the side of largest square FDGB that can be inscribed in the triangle ABE is
(a) 13/100 (b) 13/10 (c) 10/13 (d) 100/13
CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ]
Question 2
Question 3
Question 4 Important
Question 5 Important
Question 6
Question 7
Question 8 Important
Question 9 Important
Question 10
Question 11 Important
Question 12
Question 13 Important
Question 14
Question 15 Important
Question 16
Question 17 Important
Question 18
Question 19 Important
Question 20
Question 21 Important
Question 22 Important
Question 23 Important
Question 24
Question 25 Important
Question 26
Question 27
Question 28 Important
Question 29 Important
Question 30
Question 31 Important
Question 32 Important
Question 33
Question 34 Important You are here
Question 35
Question 36 Important
Question 37 Important
Question 38
Question 39 Important
Question 40
Question 41 (Case Based Question) Important
Question 42 (Case Based Question)
Question 43 (Case Based Question) Important
Question 44 (Case Based Question)
Question 45 (Case Based Question)
Question 46 (Case Based Question)
Question 47 (Case Based Question) Important
Question 48 (Case Based Question)
Question 49 (Case Based Question) Important
Question 50 (Case Based Question) Important
CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ]
Last updated at April 16, 2024 by Teachoo
Question 34 Sides AB and BE of a right triangle, right angled at B are of lengths 16 cm and 8 cm respectively. The length of the side of largest square FDGB that can be inscribed in the triangle ABE is (a) 13/100 (b) 13/10 (c) 10/13 (d) 100/13 Let FDGB be a square with side x cm Now, For parallel lines FD and CE with transversal AE ∠ ADF = ∠ AEB Also, since FDGB is a square ∠ AFD = 90° & ∠ DGF = 90° In Δ AFD and Δ DGF ∠ AFD = ∠ DGF ∠ ADF = ∠ AEB ∴ Δ AFD ~ Δ DGF Since sides in similar triangle are proportional 𝐴𝐹/𝐷𝐺=𝐹𝐷/𝐺𝐸 (𝟏𝟔 − 𝒙)/𝒙=𝒙/(𝟖 − 𝒙) (16 − x) (8 − x) = x2 16 (8 − x) − x (8 − x) = x2 128 − 16x − 8x + x2 = x2 128 − 16x − 8x = 0 128 − 24x = 0 128 = 24x 24x = 128 x = 128/24 x = 𝟏𝟔/𝟑 cm So, the correct answer is (b)