Triangles — Class 10 solved problems

33 problems from this chapter, each solved step by step.

The perimeters of two similar triangles ABC and PQR are 32 cm and 24 cm respectively. If PQ = 12 cm, find AB.
AB = 16 cm
In Δ ABC, line DE BC. If AD = 3x - 2, AE = 5x - 4, BD = 7x - 5 and CE = 5x - 3 find the value of x using the Basic Proportionality Theorem (BPT). The solution y
The value of x is 1.
In triangle ABC, DE is parallel to BC. If AD = 1.5 cm, DB = 3 cm and AE = 1 cm, find EC.
The length of EC is 2 cm.
Prove that the sum of the two acute angles of a right-angled triangle is 90 degrees.
The sum of the two acute angles of a right-angled triangle is 90^°.
Ex 6.6, 1 In Fig. 6.56, PS is the bisector of QPR of PQR. Prove that (QS)/(SR) = (PQ)/(PR). Given: PQR and PS is the bisector of QPR i.e. QPS = RPS To Prove: (Q
(QS)/(SR) = (PQ)/(PR)
In ABC, D is a point on side BC such that (BD)/(DC)=(3)/(2). If the area of ABD is 54 cm² answer the following: 1. Find the area of ACD. 2. Find the area of ABC
1. Area( ACD) = 36 cm² 2. Area( ABC) = 90 cm² 3. BE: BA = 3: 5 4. Area( BDE): Area( BAC) = 9: 25
In ABC, AD is the bisector of A, meeting side BC at D. If AB = 12 cm, AC = 18 cm, and BC = 10 cm, find the lengths of segments BD and DC.
The length of segment BD is 4 cm and the length of segment DC is 6 cm.
In ABC, AD is the bisector of A, meeting side BC at D. If AB = 10 cm, AC = 14 cm, and BC = 6 cm, find the lengths of segments BD and DC.
The length of segment BD is 2.5 cm and the length of segment DC is 3.5 cm.
In Δ ABC, AD is the bisector of A, meeting side BC at D. If AB = 10 cm, AC = 14 cm, and BC = 6 cm, find the lengths of segments BD and D
The length of segment BD is 2.5 cm and the length of segment DC is 3.5 cm.
In an obtuse-angled ABC (obtuse at B), AD is perpendicular to CB produced. Prove that AC² = AB² + BC² + 2BC × BD.
AC² = AB² + BC² + 2BC × BD
Construct a triangle ABC with AB = 5 cm, BC = 6 cm and angle B = 60° and find its area
The area of the triangle ABC is 7.5√3 cm ².
हिंदी में जवाब दो: Can a triangle have two right angles? Explain your reasoning.
No, a triangle cannot have two right angles.
In triangles ABC and PQR, AB/RQ = BC/QP = CA/PR, with angle A = 80 degrees and angle B = 60 degrees. Find angle P.
P = 40^°
In ABC, a line DE is drawn parallel to the base BC, intersecting side AB at point D and side AC at point E.If AD = x, DB = x - 2, AE = x + 2, and EC = x - 1, fi
The value of x is 4 and the total length of side AB is 6.
State and prove Basic Proportionality theorem.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same rati
In the given figure, DE || BC and all measurements are given in centimetres. The length of AE is:
The length of AE is 3 cm.
In XYZ, XY = 6 cm. If M and N are two points on XY and XZ respectively such that MN | YZ and XN = (1)/(4) XZ, then the length of XM is:
The length of XM is 1.5 cm.
In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD BC and EF AC, prove that ABD ECF.
ABD ECF
E and F are points on the sides PQ and PR respectively of a PQR and EF || QR. If PE = 2·5 cm, PQ = 6 cm and PF = 4·5 cm, find the lengths of RF and PR.
RF = 6.3 cm, PR = 10.8 cm
If a triangle has sides a = 5 cm, b = 12 cm and c = 13 cmCalculate the semi-perimeter (s)
The semi-perimeter (s) of the triangle is 15 cm.
In a △ ABC, AD is a median and AE BC. Prove that: AC²=AD²+BC· DE+((BC)/(2))²
AC²=AD²+BC· DE+((BC)/(2))²
Observe Fig. 6.30 and then find P. [Figure: ABC with AB = 3.8, BC = 6, CA = 3√3, A = 80°, B = 60°; and PQR with PQ = 12, QR = 7.6, RP = 6√3.]
P = 40^(°)
आकृति 6.31 में, OA · OB = OC · OD है। दर्शाइए कि A = C और B = D है।
चूंकि AOD COB, इसलिए A = C और B = D हैं।
यदि कोई रेखा एक ABC की भुजाओं AB और AC को क्रमश: D और E पर प्रतिच्छेद करे तथा भुजा BC के समांतर हो, तो सिद्ध कीजिए कि (AD)/(AB) = (AE)/(AC) होगा (देखिए आकृति 6.
हमने सिद्ध किया है कि यदि कोई रेखा ABC की भुजाओं AB और AC को D और E पर प्रतिच्छेद करती है और भुजा BC के समांतर है, तो (AD)/(AB) = (AE)/(AC)।
If a line intersects sides AB and AC of a ABC at D and E respectively and is parallel to BC, prove that (AD)/(AB) = (AE)/(AC) (see Fig. 6.13). [Figure: triangle
Thus, it is proven that if a line intersects sides AB and AC of a ABC at D and E respectively and is parallel to BC, then (AD)/(AB) = (AE)/(AC).
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow
The length of her shadow after 4 seconds is 1.6 m.
VARIATION: In triangle ABC, points D, E, F are on sides BC, CA, AB respectively such that AD, BE, CF are concurrent. If BD:DC = 2:3, CE:EA = 3:4, find AF:FB usi
The ratio AF:FB is 2:1.
In Fig. 6.31, OA · OB = OC · OD. Show that A = C and B = D. [Figure: segments AB and CD intersect at point O; A and D on one side, C and B on the other, forming
It has been shown that A = C and B = D using the SAS similarity criterion for triangles AOD and COB.
In triangle ABC, points D, E, F are on sides BC, CA, AB respectively such that AD, BE, CF are concurrent. If BD:DC = 2:3, CE:EA = 3:4, find AF:FB using Ceva's t
The ratio AF:FB is 2:1.
आकृति 6.16 में (PS)/(SQ) = (PT)/(TR) है तथा PST = PRQ है। सिद्ध कीजिए कि PQR एक समद्विबाहु त्रिभुज है।
PQR एक समद्विबाहु त्रिभुज है।
Q. 1: In the given figure, PS/SQ = PT/TR and ∠ PST = ∠ PRQ. Prove that PQR is an isosceles triangle. (isosceles triangle, ratio, angles, proportionality, parall
PQR is an isosceles triangle.
In Fig. 6.16, (PS)/(SQ) = (PT)/(TR) and PST = PRQ. Prove that PQR is an isosceles triangle. [Figure: triangle PQR with vertex P at top; points S on PQ and T on
Triangle PQR is an isosceles triangle.
आकृति 6.30 में P ज्ञात कीजिए। [आकृति: ABC में AB = 3.8, BC = 6, CA = 3√3, A = 80°, B = 60°; तथा PQR में PQ = 12, QR = 7.6, RP = 6√3।]
P = 40°

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