CBSE Class 10 Mathematics Triangles Assignment Set 18

Exercise-2 Objective Questions

Teacher's Note:
1. Subtract the small triangle from the big one.

Question 2. The median AD of \( \triangle ABC \) meets BC at D. The internal bisectors of \( \angle ADB \) and \( \angle ADC \) meet AB and AC at E and F respectively. Then EF
(A) is perpendicular to AD   (B) is parallel to BC   (C) divides AD in the ratio AB : AC   (D) none of these
Answer:
Ans. (B) is parallel to BC
DE bisects \( \angle ADB \): \( \dfrac{AE}{EB} = \dfrac{AD}{DB} \). DF bisects \( \angle ADC \): \( \dfrac{AF}{FC} = \dfrac{AD}{DC} \).
D is the mid-point of BC, so DB = DC, hence \( \dfrac{AE}{EB} = \dfrac{AF}{FC} \) and \( EF \parallel BC \).
In simple words: Use the angle bisector theorem twice. | DB = DC because AD is a median.

Teacher's Note:
1. Equal ratios give a parallel line.

Teacher's Note:
1. The quadrilateral is a trapezium with two vertical sides.

Teacher's Note:
1. Use similar triangles, then the triangle inequality.

Question 5. ABCD is a parallelogram, P is a point on AB such that AP : PB = 3 : 2. Q is a point on CD such that CQ : QD = 7 : 3. If PQ meets AC at R, then AR : AC is
(A) 5 : 11   (B) 6 : 13   (C) 4 : 7   (D) 2 : 5
Answer:
Ans. (B) 6 : 13
Take AB = DC = 10. AP = 6, DQ = 3, CQ = 7. Let PQ meet AC at R. Using coordinates A(0, 0), B(10, 0), D(a, h), C(10 + a, h): P = (6, 0), Q = (a + 3, h).
Point on AC: \( (t(10 + a), th) \); point on PQ: \( (6 + s(a - 3), sh) \). Equal heights give s = t, then \( 10t = 6 - 3t \Rightarrow t = \dfrac{6}{13} \). So AR : AC = 6 : 13.
In simple words: A coordinate setup works for any parallelogram. | Alternatively use similar triangles ARP and CRQ: AR : RC = AP : CQ = 6 : 7.

Teacher's Note:
1. R divides AC in the ratio 6 : 7.

Teacher's Note:
1. Angle bisector gives a 3-4-5 triangle.

Question 7. In a right triangle with sides a and b, and hypotenuse c, the altitude drawn on the hypotenuse is x. Then which one of the following is correct?
(A) \( ab = x^2 \)   (B) \( \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{x} \)   (C) \( a^2 + b^2 = 2x^2 \)   (D) \( \dfrac{1}{x^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} \)
Answer:
Ans. (D) \( \dfrac{1}{x^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} \)
Area \( = \dfrac{1}{2}ab = \dfrac{1}{2}cx \Rightarrow x = \dfrac{ab}{c} \).
\( \dfrac{1}{x^2} = \dfrac{c^2}{a^2b^2} = \dfrac{a^2 + b^2}{a^2b^2} = \dfrac{1}{b^2} + \dfrac{1}{a^2} \).
In simple words: The altitude is ab/c.

Teacher's Note:
1. Find the area two ways.

Teacher's Note:
1. Square the relation MA = h + d − x.

Question 9. A rhombus is inscribed in triangle ABC in such a way that one of its vertices is A and two of its sides lie along AB and AC, and the fourth vertex lies on BC. If AC = 6, AB = 12 and BC = 8, the side of the rhombus is
(A) 2   (B) 3   (C) 4   (D) 5
Answer:
Ans. (C) 4
Let the side be s with vertices P on AB, Q on AC, R on BC. \( PR \parallel AC \) gives \( \triangle BPR \sim \triangle BAC \): \( \dfrac{PR}{AC} = \dfrac{BP}{BA} \).
\( \dfrac{s}{6} = \dfrac{12 - s}{12} \Rightarrow 2s = 12 - s \Rightarrow s = 4 \).
In simple words: The small triangle at B is similar to ABC.

Teacher's Note:
1. The rhombus side divides AB in the ratio 4 : 8.

Question 10. ABCD (in order) is a rectangle with \( AB = CD = \dfrac{12}{5} \) and BC = DA = 5. The point P is taken on AD such that \( \angle BPC = 90^\circ \). The value of (BP + PC) is
(A) 5   (B) 6   (C) 7   (D) 8
Answer:
Ans. (C) 7
Let AP = u. \( BP^2 + PC^2 = BC^2 = 25 \): \( 2 \times \dfrac{144}{25} + u^2 + (5 - u)^2 = 25 \Rightarrow u^2 - 5u + 5.76 = 0 \Rightarrow u = 1.8 \) or 3.2.
Either way \( BP = 4 \) and \( PC = 3 \) (or the reverse), so \( BP + PC = 7 \).
In simple words: Use the Pythagoras theorem in the right triangles ABP and PDC.

Teacher's Note:
1. The 3-4-5 triangle appears.

Teacher's Note:
1. Triangle DEC has half the area of the rectangle.

Teacher's Note:
1. Each side is the mean proportional of its part of the diagonal and the diagonal.

Teacher's Note:
1. The two triangles share the ratio 2 : 3 on every side.

Question 14. O is any point inside the rectangle PQRS. Then
(A) \( OP^2 + OR^2 = OQ^2 + OS^2 \)   (B) \( OP^2 + OQ^2 = OR^2 + OS^2 \)   (C) \( OP^2 + OS^2 = OQ^2 + QR^2 \)   (D) None of the above
Answer:
Ans. (A) \( OP^2 + OR^2 = OQ^2 + OS^2 \)
Draw lines through O parallel to the sides; let the distances from O to the sides be p, q (horizontal) and r, s (vertical) with the rectangle having width p + q and height r + s.
\( OP^2 + OR^2 = (p^2 + r^2) + (q^2 + s^2) = OQ^2 + OS^2 \) (both sides equal \( p^2 + q^2 + r^2 + s^2 \)).
In simple words: This is the British flag theorem.

Teacher's Note:
1. Opposite corners give the same total of squared distances.