DAV Class 7 Maths Chapter 8 Brain Teasers
Triangle And Its Properties Brain Teasers
1. A. Tick (✔) the correct option.
(a) In \( \triangle ABC \), \( AB + BC > \) ___________. (i) \( AB \) (ii) \( BC + AC \) (iii) \( AC \) (iv) None
Solution
\( AB + BC > \color{red} AC \)
The sum of the lengths of any two sides of a triangle is greater than the length of third side.
Answer \( {\color{orange} (iii)} \ \color{red} AC \)
(b) The centroid of a triangle divides each median in the ratio— (i) 2 : 1 (ii) 2 : 3 (iii) 3 : 1 (iv) 1 : 3
Answer \( {\color{orange} (i)} \ \color{red} 2:1 \)
The centroid divides each median in the ratio \( \color{red} 2:1 \)
(c) The hypotenuse of a right triangle is 17 cm and one of the sides is 8 cm. The other side is— (i) 16 cm (ii) 12 cm (iii) 13 cm (iv) 15 cm
Solution
\begin{align*} \text{Pythagoras theorem} \\ \color{magenta} \text{(Side 1)}^2 + \text{(Side 2)}^2 &= \color{magenta} \text{(Hypotenuse)}^2 \\ 8^2 + x^2 &= 17^2 \\ 64 + x^2 &= 289 \\ x^2 &= 225 \\ x &= 15 \\ \end{align*}
Answer \( {\color{orange} (d)} \ \color{red} 15 \text{ cm} \)
(d) The point of concurrence of the altitudes of a triangle is called— (i) Orthocentre (ii) Circumcentre (iii) Centroid (iv) Incentre
Answer \( {\color{orange} (i)} \ \color{red} \text{Orthocentre} \)
The point of concurrence of the altitudes of a triangle is called the \(\color{red} orthocentre \).
(v) In \( \triangle PQR \), if \( \angle PRS \) is an exterior angle at R, then the interior adjacent angle at R is— (i) \( \angle RQP \) (ii) \( \angle QPR \) (iii) \( \angle PRQ \) (iv) \( \angle PQR \)
Solution
The adjacent interior angle to an exterior angle is the opposite interior angle.
Answer \( {\color{orange} (iii)} \ \color{red} \angle PRQ \)
B. Answer the following questions.
(a) What is the point of concurrence of the angle bisector of a triangle called? Where does it lie?
Solution
The point of concurrence of the angle bisectors of a triangle is called the incentre of the triangle. Since the angle bisectors of a triangle will all lie in the interior of the triangle, the incentre of a triangle always lies in its interior.
(b) The vertical angle of an isosceles triangle is 80°. Find the measurement of its base angles.
Solution
Let the base angles be \( x \) each. \begin{align*} 80 + x + x &= 180 \text{ (Angle sum property)}\\ 80 + 2x &= 180^\circ \\ 2x &= 180 - 80^\circ \\ 2x& = 100^\circ \\ x& = \frac{100^\circ}{2} \\ x &= 50^\circ \end{align*}
Answer Base angles are \( \color{red} 50^\circ,50^\circ \)
(c) Find the value of \( x \) and \( y \) in the given figure.
Solution
\begin{align*} x + 50^\circ &= 120^\circ \text{ (Exterior angle property)} \\ x &= 120^\circ - 50^\circ \\ x &= 70^\circ \\ \\ y + 120^\circ &= 180^\circ \text{ (Linear pair)} \\ y &= 180^\circ - 120^\circ \\ y &= 60^\circ \end{align*}
Answer \( \color{red} x = 70^\circ, \ y = 60^\circ \)
(d) Can the line segments 1.5 cm, 2 cm, and 2.5 cm be the sides of a right-angled triangle?
Solution
\begin{align*} \text{Longest side} & = 2.5 \ cm \\ \text{Side 1} & = 2 \ cm \\ \text{Side 2} & = 1.5 \ cm \\ \text{Pythagoras theorem} \\ \color{magenta} \text{(Side 1)}^2 + \text{(Side 2)}^2 &= \color{magenta} \text{(Hypotenuse)}^2 \\ 2^2 + (1.5)^2 &= (2.5)^2 \\ 4 + 2.25 &= 6.25 \\ 6.25 &= 6.25 \end{align*}
Answer \( \color{red} \text{Yes} \), the given are sides of a right-angled triangle.
(e) Where does the circumcentre of a right-angled triangle lie?
Answer In a right angled triangle, the circumcentre is at the mid-point of the hypotenuse.
2. A man goes 5 m in east and then 12 m in north direction. Find the distance from the starting point.
Solution
\begin{align*} AC & = x \\ AB & = 5 \ m \\ BC & = 12 \ m \\ &\text{Pythagoras theorem} \\ \color{magenta} (AC)^2 &= \color{magenta} (AB)^2 + (BC)^2 \\ x^2 &= (5)^2 + (12)^2 \\ x^2 &= 25 + 144 \\ x^2 &= 169 \\ x &= 13 \ m \\ \end{align*}
Answer Distance from the starting point \( \color{red} 13 \ m \).
3. The width of a rectangle is 5 cm and its diagonal is 13 cm. What is its perimeter?
Solution
\begin{align*} In \ & \triangle \ ABC \\ AC & = 13 \ cm \\ BC & = 5 \ cm \\ AB & = x \\ \text{Pythagoras theorem} \\ \color{magenta} (AB)^2 + (BC)^2 & = \color{magenta} (AC)^2 \\ x^2 + (5)^2 &= (13)^2 \\ x^2 + 25 &= 169 \\ x^2 &= 169 - 25 \\ x^2 &= 144 \\ x &= 12 \ m \\ AB &= 12 \ m \\ \\ Perimeter & = 2(L+B) \\ & = 2(12+5) \\ & = 2 \times 17 \\ Perimeter & = 34 \ cm \end{align*}
Answer Perimeter \( \color{red} 34 \ cm \).
4. In the given figure \( \triangle PQR \) and \( \triangle SQR \) are isosceles.
(i) Find \( \angle PQR \) and \( \angle PRQ \)
Solution
\begin{align*} In \ & \triangle \ PQR \\ PQ &= PR \\ \text{Angles opposite to equal }& \text{sides of a triangle are equal}\\ \angle PQR &= \angle PRQ \\ \text{By the angle sum }& \text{property}\\ \angle P + \angle PQR + \angle PRQ &= 180^\circ \\ 30^\circ + \angle PQR + \angle PQR &= 180^\circ \\ 2 \angle PQR &= 180^\circ \\ 2 \angle PQR &= 180^\circ - 30^\circ \\ 2 \angle PQR &= 150^\circ \\ \angle PQR &= \frac{150^\circ}{2} \\ \angle PQR &= 75^\circ \\ \angle PRQ &= 75^\circ \end{align*}
Answer \( \angle PQR = {\color{red} 75^\circ} , \angle PRQ = {\color{red} 75^\circ}\).
(ii) Find \( \angle SQR \) and \( \angle SRQ \)
Solution
\begin{align*} In \ & \triangle \ SQR \\ SQ &= SR \\ \text{Angles opposite to equal }& \text{sides of a triangle are equal}\\ \angle SQR &= \angle SRQ \\ \text{By the angle sum }& \text{property}\\ \angle S + \angle SQR + \angle SRQ &= 180^\circ \\ 70^\circ + \angle SQR + \angle SQR &= 180^\circ \\ 2 \angle SQR &= 180^\circ \\ 2 \angle SQR &= 180^\circ - 70^\circ \\ 2 \angle SQR &= 110^\circ \\ \angle SQR &= \frac{110^\circ}{2} \\ \angle SQR &= 55^\circ \\ \angle SRQ &= 55^\circ \end{align*}
Answer \( \angle SQR = {\color{red} 55^\circ} , \angle SRQ = {\color{red} 55^\circ}\).
(iii) Find \( x \)
Solution
\begin{align*} x & = \angle PQR - \angle SRQ \\ x & = 75^\circ - 55^\circ \\ x & = 20^\circ \\ \end{align*}
Answer \( x = {\color{red} 20^\circ}\)
5. In the given figure \(\triangle PQR\) is isosceles with \(PQ = PR \). \( LM\) is parallel to \(QR\) with \(L\) on \(PQ\) and \(M\) on \(PR\). Give reasons for each of the following statements:
(i) \(\angle Q = \angle R\)
Answer Angles opposite to equal sides of a triangle are equal
\begin{align*} & In \ \triangle \ PQR \\ & PQ = PR \ (Given) \\ & \color{red} \text{Angles opposite to equal sides of a triangle are equal}\\ & \implies \angle Q = \angle R \end{align*}
(ii) \(\angle PLM = \angle Q \)
Answer Corresponding angles
\begin{align*} & In \ \triangle \ PQR \\ & LM \parallel QR \ (Given) \\ & PQ \ \text{(Transversal line)} \\ & \implies \angle PLM = \angle Q \color{red} \text{ (Corresponding angles)} \end{align*}
(iii) \(\angle PML = \angle R \)
Answer Corresponding angles
\begin{align*} & In \ \triangle \ PQR \\ & LM \parallel QR \ (Given) \\ & PR \ \text{(Transversal line)} \\ & \implies \angle PML = \angle R \color{red} \text{ (Corresponding angles)} \end{align*}
(iv) \( \angle PLM = \angle PML \)
Answer From (i) , (ii) and (iii)
\begin{align*} \angle Q &= \angle R \quad (i) \\ \angle PLM &= \angle Q \quad (ii) \\ \angle PML &= \angle R \quad (iii) \\ \implies \angle PLM &= \angle PML \end{align*}
(v) \( \triangle PLM \) is isosceles.
Answer In a triangle, if two angles are equal, then the sides opposite to these angles are also equal.
\begin{align*} & In \ \triangle \ PLM \\ & \angle PLM = \angle PML \implies \text{from (iv)} \\ \end{align*}
In a triangle, if two angles are equal, then the sides opposite to these angles are also equal.
\begin{align*} & PL = PM \\ & \triangle PLM \color{red} \text{ is isosceles} \end{align*}6. In an isosceles triangle, the base angle is twice as large as vertex angle. Find the angles of the triangle.
Solution
\begin{align*} \text{Let the vertex angle be } & = x \\ \text{Let base angles be } & = 2x \\ \text{(Angle sum } & \text{property)} \\ \color{magenta} \angle A + \angle B + \angle C & = \color{magenta} 180^\circ \\ 2x + 2x + x &= 180^\circ \\ 5x &= 180^\circ \\ x &= 36^\circ \\ \color{green} \text{Vertex angle} & = \color{green} 36^\circ \\ \text{Base angles} & = 2x\\ & = 2 \times 36^\circ\\ \color{green} \text{Base angles} & = \color{green} 72^\circ\\ \end{align*}
Answer Angles of the triangle are \( \color{red} 36^\circ , 72^\circ, 72^\circ \).
7. A pole broke at a point but did not separate. Its top touched the ground at a distance of 5m from its base. If the point where it broke is at a height of 12 m from the ground, what was the total height of the pole before it broke?
Solution
\begin{align*} AC & = x \\ AB & = 5 \ m \\ BC & = 12 \ m \\ &\text{Pythagoras theorem} \\ \color{magenta} (AC)^2 &= \color{magenta} (AB)^2 + (BC)^2 \\ x^2 &= (5)^2 + (12)^2 \\ x^2 &= 25 + 144 \\ x^2 &= 169 \\ x &= 13 \ m \\ \\ &\text{Total height of the pole} \\ & = BC + CA \\ & = 12 \ m + 13 \ m \\ & = 25 \ m \\ \end{align*}
Answer Total height of the pole \( \color{red} 25 \ m \)
8. ∆ ABC is right angled at C. Can you locate its orthocentre without drawing any altitude? If so, name it.
Answer Yes, it is the vertex C.
9. Draw an equilateral triangle each of whose sides is 6 cm. Draw its medians. Are they equal?
Answer All the three medians are equal 5.2cm
10. The exterior angle at C of a \(\triangle ABC\) is equal to \(120^\circ\). Find the measure of all the angles of the triangle if \(\angle A = \angle B\). What type of triangle is this?
Solution
\begin{align*} \text{Given :} \\ \angle ACD &= 120^\circ \\ \angle A &= \angle B \\ \\ \text{To find} & = \angle A \ , \angle B \ , \angle ACB \\ \\ \color{magenta} \angle ACB + \angle ACD & = \color{magenta} 180^\circ \text{ (Linear pair)} \\ \angle ACB + 120^\circ & = 180^\circ \\ \angle ACB & = 180^\circ - 120^\circ \\ \color{green} \angle ACB & = \color{green} 60^\circ \\ \\ \color{magenta}\angle A + \angle B &= \color{magenta} \angle ACD \text{ (Exterior angle property)} \\ \angle A + \angle A &= 120^\circ \\ 2 \ \angle A &= 120^\circ \\ \angle A &= \frac{120^\circ}{2} \\ \\ \angle A &= 60^\circ \\ \therefore \angle B &= 60^\circ \\ \angle ACB & = 60^\circ \\ \triangle ABC & \text{ is an equilateral triangle} \end{align*}
Answer The angles of the triangle are \(\color{red}{60^\circ,\, 60^\circ,\, 60^\circ.}\) It is an \(\color{red}{\text{equilateral triangle.}}\)
11. In the given figure, find the measure of \(\angle 1 , \angle 2, \angle 3\)
Solution
\begin{align*} \angle 3 + 136^\circ & = 180^\circ \color{magenta} \text{ (Linear pair)} \\ \angle 3 & = 180^\circ - 136^\circ \\ \color{green}\angle 3 & = \color{green}44^\circ \\ \\\angle 2 + 104^\circ & = 180^\circ \color{magenta} \text{ (Linear pair)} \\ \angle 2 & = 180^\circ - 104^\circ \\ \color{green}\angle 2 & = \color{green}76^\circ \\ \\\angle 1 + \angle 2 + \angle 3 & = 180^\circ \color{magenta} \text{ (Angle sum property)} \\ \angle 1 + 76^\circ + 44^\circ & = 180^\circ \\ \angle 1 + 120^\circ & = 180^\circ \\ \angle 1 & = 180^\circ - 120^\circ \\ \color{green} \angle 1 & = \color{green} 60^\circ \end{align*}
Answer \( \angle 1 = {\color{red} 60^\circ} , \angle 2 = {\color{red} 76^\circ} ,\angle 3 = {\color{red} 44^\circ} \)
12. In the given figure, find the measure of \(\angle CAB , \angle ABC, \angle ACB \)
Solution
\begin{align*} \angle CAB & = \angle EAF \color{magenta} \text{ (Vertically opposite angles)} \\ \color{green}\angle CAB & = \color{green} 45^\circ \\ \\ \angle ABC + \angle CAB & = \angle ACD \color{magenta} \text{ (Exterior angle property)} \\ \angle ABC + 45^\circ & = 105^\circ \\ \angle ABC & = 105^\circ - 45^\circ \\ \color{green} \angle ABC & = \color{green} 60^\circ \\ \\\angle ACB + \angle ACD & = 180^\circ \color{magenta} \text{ (Linear pair)} \\ \angle ACB + 105^\circ & = 180^\circ \\ \angle ACB & = 180^\circ - 105^\circ \\ \color{green}\angle ACB & = \color{green} 75^\circ \end{align*}
Answer \( \angle CAB = {\color{red} 45^\circ} , \angle ABC = {\color{red} 60^\circ} ,\angle ACB = {\color{red} 75^\circ} \)
13. Find x and y in the following figure.
Solution
\begin{align*} \angle x &= \angle A + \angle B \color{magenta} \text{ (Exterior angle property)} \\ x &= 30^\circ + 30^\circ \\ \color{green} x &= \color{green} 60^\circ \\ \\ \angle y &= \angle D + \angle DCE \\ y &= 30^\circ + 60^\circ \\ \color{green} y &= \color{green} 90^\circ \\ \end{align*}
Answer \( x = {\color{red} 60^\circ} , y = {\color{red} 90^\circ} \)
14. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Solution
The sum of any two sides of a triangle is always greater than the third side.Length of the third side is greater than 15 - 12 = 3 cm.
Length of the third side is less than 15 + 12 = 27 cm.
Answer \( 3 cm < {\color{red} \text{Length of the third side}} < 27cm \)