Download Class 10 Mathematics Concept Summaries: CBSE Class 10 Mathematics Triangles Notes Set 09
Access comprehensive revision notes for Chapter 06 Triangles using the CBSE Class 10 Mathematics Triangles Notes Set 09. Designed to align with the 2026-27 academic syllabus for Class 10 Mathematics, these concept summaries help students streamline their exam preparation and review complex topics efficiently.
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Areas of Similar Triangles
Statement. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Given. Two triangles ABC and PQR such that \( \triangle ABC \sim \triangle PQR \).
To prove. \( \dfrac{ar(ABC)}{ar(PQR)} = \left(\dfrac{AB}{PQ}\right)^2 = \left(\dfrac{BC}{QR}\right)^2 = \left(\dfrac{CA}{RP}\right)^2 \).
Construction. Draw the altitudes AM and PN of \( \triangle ABC \) and \( \triangle PQR \).
Proof. \( ar(ABC) = \tfrac{1}{2} BC \times AM \) and \( ar(PQR) = \tfrac{1}{2} QR \times PN \). So
\[ \frac{ar(ABC)}{ar(PQR)} = \frac{\tfrac{1}{2} BC \times AM}{\tfrac{1}{2} QR \times PN} = \frac{BC \times AM}{QR \times PN} \qquad \text{... (i)} \]
In \( \triangle ABM \) and \( \triangle PQN \): \( \angle B = \angle Q \) (as \( \triangle ABC \sim \triangle PQR \)) and \( \angle M = \angle N = 90^\circ \). So \( \triangle ABM \sim \triangle PQN \) (AA), and \( \dfrac{AM}{PN} = \dfrac{AB}{PQ} \) ... (ii). Also \( \triangle ABC \sim \triangle PQR \) gives \( \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP} \) ... (iii). Therefore, from (i) and (ii), \( \dfrac{ar(ABC)}{ar(PQR)} = \dfrac{BC}{QR} \times \dfrac{AB}{PQ} \), and using (iii) this equals \( \dfrac{AB}{PQ} \times \dfrac{AB}{PQ} = \left(\dfrac{AB}{PQ}\right)^2 \). Now using (iii),
\[ \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2. \]
Properties of Areas of Similar Triangles
1. The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.
2. The areas of two similar triangles are in the ratio of the squares of the corresponding medians.
3. The areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments.
Solved Examples
Answer:
Given. A square ABCD; equilateral triangles BCE and ACF are described on the side BC and the diagonal AC. To prove. \( ar(\triangle BCE) = \tfrac{1}{2} ar(\triangle ACF) \).
Since \( \triangle BCE \) and \( \triangle ACF \) are equilateral, they are equiangular (each angle 60 degrees) and hence similar. So \( \dfrac{ar(BCE)}{ar(ACF)} = \dfrac{BC^2}{AC^2} \).
For the square, the diagonal \( AC = \sqrt{2}\,BC \), so \( \dfrac{BC^2}{AC^2} = \dfrac{BC^2}{(\sqrt{2}BC)^2} = \dfrac{1}{2} \). Hence \( ar(\triangle BCE) = \dfrac{1}{2} ar(\triangle ACF) \).
Answer:
In \( \triangle RQP \) and \( \triangle RST \): \( \angle Q = \angle S = 90^\circ \) and \( \angle PRQ = \angle SRT \) (vertically opposite angles). By AA, \( \triangle RQP \sim \triangle RST \).
\( \dfrac{RQ}{RS} = \dfrac{QP}{ST} \Rightarrow \dfrac{3}{RS} = \dfrac{4}{12} \Rightarrow RS = 9 \) m.
Area of \( \triangle RST = \dfrac{1}{2} \times 9 \times 12 = 54 \) m². Answer: (C).
Check Your Progress
1. If \( \triangle ABC \sim \triangle DEF \) such that the area of \( \triangle ABC \) is 9 cm², the area of \( \triangle DEF \) is 16 cm² and BC = 2.1 cm, find the length of EF.
Answer:
\( \dfrac{ar(ABC)}{ar(DEF)} = \dfrac{BC^2}{EF^2} \Rightarrow \dfrac{9}{16} = \dfrac{2.1^2}{EF^2} \Rightarrow \dfrac{3}{4} = \dfrac{2.1}{EF} \). EF = 2.8 cm.
Answer:
\( \triangle ALM \sim \triangle ABC \) with \( \dfrac{AL}{AB} = \dfrac{1}{4} \). So \( LM = \dfrac{1}{4} \times 8 = \) 2 cm.
\( \dfrac{ar(ABC)}{ar(ALM)} = \left(\dfrac{4}{1}\right)^2 = 16 \), so \( ar(ABC) = 16 \times 18 = \) 288 sq cm.
3. D and E are points on AB and AC of triangle ABC such that DE is parallel to BC. If AD = 3 cm, DB = 2 cm and the area of \( \triangle ABC \) is 10 sq cm, find the area of \( \triangle ADE \).
Answer:
\( \triangle ADE \sim \triangle ABC \) with \( \dfrac{AD}{AB} = \dfrac{3}{5} \). \( ar(ADE) = \dfrac{9}{25} \times 10 = \) 3.6 sq cm.
Answer:
\( \dfrac{ar(DEF)}{ar(ABC)} = \dfrac{EF^2}{BC^2} = \dfrac{16}{9} \). \( ar(DEF) = 54 \times \dfrac{16}{9} = \) 96 sq cm.
Answer:
\( \triangle ADE \) and \( \triangle CED \) have the same height from D to AC, so their areas are in the ratio of their bases AE : EC. By BPT, \( \dfrac{AE}{EC} = \dfrac{AD}{DB} = \dfrac{5}{4} \). The ratio is 5 : 4.
Pythagoras Theorem
Statement. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Given. A right triangle ABC, right angled at B.
To prove. \( AC^2 = AB^2 + BC^2 \).
Construction. \( BD \perp AC \).
Proof. In \( \triangle ADB \) and \( \triangle ABC \): \( \angle DAB = \angle CAB \) (common) and \( \angle BDA = \angle CBA = 90^\circ \). So \( \triangle ADB \sim \triangle ABC \) (AA), and \( \dfrac{AD}{AB} = \dfrac{AB}{AC} \), or \( AD \cdot AC = AB^2 \) ... (i). Similarly \( \triangle BDC \sim \triangle ABC \), so \( \dfrac{CD}{BC} = \dfrac{BC}{AC} \), or \( CD \cdot AC = BC^2 \) ... (ii). Adding (i) and (ii): \( AD \cdot AC + CD \cdot AC = AB^2 + BC^2 \), that is \( AC(AD + CD) = AB^2 + BC^2 \), or \( AC \cdot AC = AB^2 + BC^2 \). Hence \( AC^2 = AB^2 + BC^2 \).
Converse of Pythagoras Theorem
Statement. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.
Given. A triangle ABC such that \( AC^2 = AB^2 + BC^2 \).
Construction. Construct a triangle DEF such that DE = AB, EF = BC and \( \angle E = 90^\circ \).
Proof. To prove that \( \angle B = 90^\circ \), it is sufficient to show \( \triangle ABC \cong \triangle DEF \). Since \( \triangle DEF \) is right angled at E, \( DF^2 = DE^2 + EF^2 = AB^2 + BC^2 \) (by construction). But \( AB^2 + BC^2 = AC^2 \) (given), so \( DF^2 = AC^2 \), that is DF = AC ... (i). In \( \triangle ABC \) and \( \triangle DEF \): AB = DE, BC = EF (construction) and AC = DF (from (i)). So \( \triangle ABC \cong \triangle DEF \) (SSS). Hence \( \angle B = \angle E = 90^\circ \); \( \triangle ABC \) is a right triangle, right angled at B.
Some Results Deduced from Pythagoras Theorem
(i) In the given figure, \( \triangle ABC \) is obtuse angled at B. If \( AD \perp CB \) (produced), then \( AC^2 = AB^2 + BC^2 + 2BC \cdot BD \).
(ii) In the given figure, if \( \angle B \) of \( \triangle ABC \) is an acute angle and \( AD \perp BC \), then \( AC^2 = AB^2 + BC^2 - 2BC \cdot BD \).
(iii) In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side (called Apollonius Theorem).
(iv) Three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle.
Solved Examples
Answer:
(i) \( \triangle ABC \) is equilateral. In \( \triangle ABD \) and \( \triangle ACD \): AD = AD (common), \( \angle ADB = \angle ADC = 90^\circ \) and AB = AC. By RHS congruence, \( \triangle ABD \cong \triangle ACD \), so BD = DC = a.
\( \triangle ABD \) is right angled, so \( AD = \sqrt{AB^2 - BD^2} = \sqrt{4a^2 - a^2} = a\sqrt{3} \).
(ii) Area \( = \dfrac{1}{2} \times BC \times AD = \dfrac{1}{2} \times 2a \times a\sqrt{3} = \sqrt{3}\,a^2 \).
Answer:
In \( \triangle BAL \): \( BL^2 = AL^2 + AB^2 \) ... (i). In \( \triangle CAM \): \( CM^2 = AM^2 + AC^2 \) ... (ii).
Adding and multiplying by 4: \( 4(BL^2 + CM^2) = 4(AL^2 + AB^2 + AM^2 + AC^2) = 4\{AL^2 + AM^2 + (AB^2 + AC^2)\} = 4(AL^2 + AM^2 + BC^2) \) (since \( \triangle ABC \) is right angled at A) \( = 4(ML^2 + BC^2) \) (since \( \triangle LAM \) is right angled) \( = 4ML^2 + 4BC^2 \).
M and L are the midpoints of AB and AC, so \( ML = \dfrac{BC}{2} \) and \( 4ML^2 = BC^2 \). Hence \( 4(BL^2 + CM^2) = BC^2 + 4BC^2 = 5BC^2 \).
Answer:
Through O draw \( PQ \parallel BC \) so that P lies on AB and Q lies on DC. Then \( PQ \perp AB \) and \( PQ \perp DC \), so BPQC and APQD are rectangles; BP = CQ and DQ = AP.
From \( \triangle OPB \): \( OB^2 = BP^2 + OP^2 \) ... (i). From \( \triangle ODQ \): \( OD^2 = OQ^2 + DQ^2 \) ... (ii). From \( \triangle OQC \): \( OC^2 = OQ^2 + CQ^2 \) ... (iii). From \( \triangle OAP \): \( OA^2 = AP^2 + OP^2 \) ... (iv).
Adding (i) and (ii): \( OB^2 + OD^2 = BP^2 + OP^2 + OQ^2 + DQ^2 = CQ^2 + OP^2 + OQ^2 + AP^2 = (CQ^2 + OQ^2) + (OP^2 + AP^2) = OC^2 + OA^2 \) [from (iii) and (iv)]. Hence proved.
Answer:
(i) Let \( CD \perp AB \), so CD = p. Area of \( \triangle ABC = \dfrac{1}{2}(AB \times CD) = \dfrac{1}{2}cp \). Also area \( = \dfrac{1}{2}(BC \times AC) = \dfrac{1}{2}ab \). So \( cp = ab \).
(ii) Since \( \triangle ABC \) is right angled at C, \( c^2 = a^2 + b^2 \). Using \( c = \dfrac{ab}{p} \): \( \dfrac{a^2b^2}{p^2} = a^2 + b^2 \Rightarrow \dfrac{1}{p^2} = \dfrac{a^2 + b^2}{a^2b^2} = \dfrac{1}{b^2} + \dfrac{1}{a^2} \).
Answer:
Draw \( AE \perp BC \). In an equilateral triangle the altitude bisects the opposite side, so \( BE = EC = \dfrac{BC}{2} \). Since \( BD = \dfrac{BC}{3} \), \( DE = BE - BD = \dfrac{BC}{2} - \dfrac{BC}{3} = \dfrac{BC}{6} \).
In \( \triangle ABE \): \( AB^2 = AE^2 + EB^2 \) ... (i). In \( \triangle ADE \): \( AD^2 = AE^2 + ED^2 \) ... (ii). Subtracting, \( AB^2 = AD^2 - ED^2 + EB^2 \).
\( AB^2 = AD^2 - \dfrac{BC^2}{36} + \dfrac{BC^2}{4} \). With AB = BC: \( AB^2 + \dfrac{AB^2}{36} - \dfrac{AB^2}{4} = AD^2 \Rightarrow \dfrac{36AB^2 + AB^2 - 9AB^2}{36} = AD^2 \Rightarrow \dfrac{28AB^2}{36} = AD^2 \Rightarrow 7AB^2 = 9AD^2 \).
Check Your Progress
1. A man of height 1.8 m, standing 5 m away from a lamp post, observes that the length of his shadow is 1.5 m. What is the height of the lamp post?
Answer:
The man and the lamp post form similar right triangles. The tip of the shadow is 5 + 1.5 = 6.5 m from the post. \( \dfrac{H}{1.8} = \dfrac{6.5}{1.5} \Rightarrow H = \) 7.8 m.
2. A rope from the top of a mast on a sailboat is attached to a point 2 metres from the base of the mast. The rope is 8 metres long. How high is the mast?
Answer:
Height \( = \sqrt{8^2 - 2^2} = \sqrt{60} = \) \( 2\sqrt{15} \) m.
3. Identify the triangle as acute angled, obtuse angled or right angled, whose sides are given: (i) a = 12, b = 15, c = 20 (ii) a = 15, b = 8, c = 17 (iii) a = 12, b = 5, c = 17 (iv) a = 8, b = 9, c = 12.
Answer:
Compare the square of the longest side with the sum of the squares of the other two.
(i) \( 144 + 225 = 369 < 400 \): obtuse. (ii) \( 225 + 64 = 289 = 17^2 \): right. (iii) \( 144 + 25 = 169 < 289 \): obtuse. (iv) \( 64 + 81 = 145 > 144 \): acute.
4. In a triangle ABC, \( \angle A = 90^\circ \). If \( AD \perp BC \), prove that \( AB^2 - BD^2 = AC^2 - CD^2 \).
Answer:
In the right triangles ADB and ADC, \( AB^2 - BD^2 = AD^2 \) and \( AC^2 - CD^2 = AD^2 \). Hence \( AB^2 - BD^2 = AC^2 - CD^2 \).
5. A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time another tower casts a shadow 40 m long on the ground. Find the height of the tower.
Answer:
The triangles are similar: \( \dfrac{H}{12} = \dfrac{40}{8} \Rightarrow H = \) 60 m.
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Revision Notes and Key Concepts for Class 10 Mathematics Chapter 06 Triangles
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