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Maths 5 Mark Question

Understanding Shapes - II (Quadrilaterals) · Maharashtra Board STD 8 (MSBSHSE - Semi English Medium). 8 questions.

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Question 15 Marks
The four angles of a quadrilateral are as 3 : 5 : 7 : 9. Find the angles.
Answer
Sum of four angles of a quadrilateral = 360° and ratio in angles = 3 : 5 : 7 : 9 Let first angles = 2x Then second angle = 5x third angle = 7x and fourth angle = 9x 3x + 5x + 7x + 9x = 360° ⇒ 24x = 369° $\Rightarrow\text{x}=\frac{360}{24}=15^\circ$First angle
= 3x = 3 × 15° = 45° Second angle = 5x = 5 × 15° = 75° Third angle = 7x = 7 × 15° = 105° Fourth angle = 9x = 9 × 15° = 135°
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Question 25 Marks
In a convex hexagon, prove that the sum of all interior angles is equal to twice the sum of its exterior angles formed by producing the sides in the same order.
Answer

In a convex hexagon ABCDEF, its sides AB, BG, CD, DE, EF and FA are produced in order forming exterior angles a, b, c, d, e, f
$\angle\text{a}+\angle\text{b}+\angle\text{c}+\angle\text{d}+\angle\text{e}+\angle\text{f}=4$ right angles (By definition)
By joining AC, AD, and AE, 4 triangles ABC, ACD, ADE and AEF are formed
$\text{In}\ \triangle\text{ABC}$,
$\angle1+\angle2+\angle3=180^\circ=2$ right angle (Sum of angles of a triangle) …… (i)
Similarly,
$\text{In}\ \triangle\text{ACD}$,
$\angle4+\angle5+\angle6=180^\circ=2$ right angles
$\text{In}\ \triangle\text{ADE}$,
$\angle1+\angle8+\angle9=2$ right angles …(iii)
$\text{In}\ \triangle\text{AEF}$,
$\angle10+\angle11+\angle12=2$ right angles …(iv)
Joining (i), (ii), (iii) and (iv)
$\angle1+\angle2+\angle3+\angle4+\angle5+\angle6+\\\angle7+\angle8+\angle9+\angle10+\angle11+\angle12=8$ right angles
$\Rightarrow\angle2+\angle3+\angle5+\angle6+\angle8+\angle9+\\\angle11+\angle12+\angle1+\angle4+\angle7+\angle10=8$ right angles
$\Rightarrow\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{f}+\angle\text{A}=8$ right angles
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{f}=2$
$\Big(\angle\text{a}+\angle\text{b}+\angle\text{c}+\angle\text{d}+\angle\text{e}+\angle\text{f}\Big)$
Sum of all interior angles = 2(The sum of exterior angles)
Hence proved.
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Question 35 Marks
Determine the number of sides of a polygon whose exterior and interior angles are in the ratio 1 : 5.
Answer
Ratio in exterior angle and interior angles of a regular polygon = 1 : 5
But sum of interior and exterior angles = 180° (Linear pair)
$\therefore\text{Exterior angle}=\frac{180^\circ\times1}{1+5}$
$=\frac{180^\circ\times1}{6}=30^\circ$
$=\frac{180^\circ\times1}{6}=30^\circ$
and Interior angles $=\frac{180^\circ\times5}{6}=150^\circ$
Let number of sides of the polygon = n
$\therefore\frac{2\text{n}-4}{\text{n}}\times90^\circ=150^\circ$
$\Rightarrow\frac{2\text{n}-4}{\text{n}}=\frac{150}{90}=\frac{5}{3}$
By Cross multiplication:
6n – 12 = 5n
⇒ 6n – 5n = 12
⇒ n = 12
Number of sides of polygon is 12.
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Question 45 Marks
In the figure, the bisectors of $\angle\text{A}$ and $\angle\text{B}$ meet at a point P. If $\angle\text{C}=100^\circ$ and $\angle\text{D}=50^\circ$, find the measure of $\angle\text{APB}$.
Answer

In quadrilateral ABCD,
$\angle\text{D}=50^\circ$
$\angle\text{C}=100^\circ$
PA and PB are the bisectors of $\angle\text{A}$ and $\angle\text{B}$.
In quadrilateral ABCD,
$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$ (Sum of angles of a quadrilateral)
$\Rightarrow\angle\text{A}+\angle\text{B}+100^\circ+50^\circ=360^\circ$
$\Rightarrow\angle\text{A}+\angle\text{B}+150^\circ=360^\circ$
$\Rightarrow\angle\text{A}+\angle\text{B}=360^\circ-150^\circ=210^\circ$
and
$\frac{1}{2}+\angle\text{A}+\frac{1}{2}+\angle\text{B}=\frac{210}{2}$
$=105^\circ$
(PA and PB are bisector of $\angle\text{A}$ and $\angle\text{B}$ respectively)
$\angle\text{PAB}+\angle\text{PBA}=105^\circ$
$\Rightarrow\angle\text{PAB}+\angle\text{PBA}+\angle\text{APB}=180^\circ$ (Sum of angles of a triangle)
$\Rightarrow105^\circ+\angle\text{APB}=180^\circ$
$\Rightarrow\angle\text{APB}=180^\circ-105^\circ=75^\circ$
$\angle\text{APB}=75^\circ$
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Question 55 Marks
In a convex hexagon, prove that the sum of all interior angles is equal to twice the sum of its exterior angles formed by producing the sides in the same order.
Answer

In a convex hexagon ABCDEF, its sides AB, BG, CD, DE, EF and FA are produced in order forming exterior angles a, b, c, d, e, f
$\angle\text{a}+\angle\text{b}+\angle\text{c}+\angle\text{d}+\angle\text{e}+\angle\text{f}=4$ right angles (By definition)
By joining AC, AD, and AE, 4 triangles ABC, ACD, ADE and AEF are formed
$\text{In}\ \triangle\text{ABC}$,
$\angle1+\angle2+\angle3=180^\circ=2$ right angle (Sum of angles of a triangle) …… (i)
Similarly,
$\text{In}\ \triangle\text{ACD}$,
$\angle4+\angle5+\angle6=180^\circ=2$ right angles
$\text{In}\ \triangle\text{ADE}$,
$\angle1+\angle8+\angle9=2$ right angles …(iii)
$\text{In}\ \triangle\text{AEF}$,
$\angle10+\angle11+\angle12=2$ right angles …(iv)
Joining (i), (ii), (iii) and (iv)
$\angle1+\angle2+\angle3+\angle4+\angle5+\angle6+\\\angle7+\angle8+\angle9+\angle10+\angle11+\angle12=8$ right angles
$\Rightarrow\angle2+\angle3+\angle5+\angle6+\angle8+\angle9+\\\angle11+\angle12+\angle1+\angle4+\angle7+\angle10=8$ right angles
$\Rightarrow\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{f}+\angle\text{A}=8$ right angles
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{f}=2$
$\Big(\angle\text{a}+\angle\text{b}+\angle\text{c}+\angle\text{d}+\angle\text{e}+\angle\text{f}\Big)$
Sum of all interior angles = 2(The sum of exterior angles)
Hence proved.
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Question 65 Marks
In a quadrilateral ABCD, CO and DO are the bisectors of $\angle\text{C}$ and $\angle\text{D}$ respectively. Prove that $\angle\text{COD}=\frac{1}{2}(\angle\text{A}+\angle\text{B})$.
Answer

In quadrilateral ABCD,
$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
(Sum of angle of a quadrilateral)
$\Rightarrow\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\frac{1}{2}\angle\text{D}=180^\circ$
(Dividing by 2)....(i)
But in $\triangle\text{COD},$
$\frac{1}{2}\angle\text{C}+\frac{1}{2}\angle\text{D}+\angle\text{COD}=180^\circ$
(Sum of angle of a triangle).......(ii)
From (i) and (ii),
$\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\frac{1}{2}\angle\text{D}\\=\frac{1}{2}\angle\text{C}+\frac{1}{2}\angle\text{D}+\angle\text{COD}$
$\Rightarrow\angle\text{COD}=\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}$
$\Rightarrow\angle\text{COD}=\frac{1}{2}\big(\angle\text{A}+\angle\text{B}\big)$
Hence proved
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Question 75 Marks
PQRSTU is a regular hexagon. Determine each angle of $\triangle\text{PQT}$.
Answer

$\therefore$ If each interior angle $=\frac{2\text{n}-4}{\text{n}}\times90^\circ$
$=\frac{2\times6-4}{\text{6}}\times90^\circ$
$=\frac{8}{\text{6}}\times90^\circ=120^\circ$
In $\triangle\text{PUT},\text{PU}=\text{UT}$
$\angle\text{UPT}=\angle\text{UTP}$
But $\angle\text{UPT}+\angle\text{UTP}=180^\circ-\angle\text{U}$
$=180^\circ-120^\circ=60^\circ$
$\angle\text{UPT}=\angle\text{UTP}=30^\circ$
$\angle\text{TPQ}=120^\circ-30^\circ=90^\circ$ (QT is diagonal which bisect ∠Q and ∠T)
$\angle\text{PQT}=\frac{120}{2}=60^\circ$
Now in $\triangle\text{PQT}$,
$\angle\text{TPQ}+\angle\text{PQT}+\angle\text{PTQ}=180^\circ$ (Sum of angles of a triangle)
$\Rightarrow90^\circ+60^\circ+\angle\text{PTQ}=180^\circ$
$\Rightarrow150^\circ+\angle\text{PTQ}=180^\circ$
$\Rightarrow\angle\text{PTQ}=180^\circ-150^\circ=30^\circ$
Hence in $\triangle\text{PQT}$,
$\angle\text{P}=90^\circ$
$\angle\text{Q}=60^\circ$
and
$\angle\text{T}=30^\circ$
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Question 85 Marks

In the figure, ABCD is a quadrilateral.
  1. Name a pair of adjacent sides.
  2. Name a pair of opposite sides.
  3. How many pairs of adjacent sides are there?
  4. How many pairs of Opposite sides are there?
  5. Name a pair of adjacent angles.
  6. Name a pair of opposite angles.
  7. How many pairs of adjacent angles are there?
  8. How many pairs of opposite angles are there?
Answer
In the figure, ABCD is a quadrilateral,
  1. Pairs of adjacent sides are AB, BC, BC, CD, CD, DA, DA, AB.
  2. Pairs of opposite sides are AB and CD, BC and AD.
  3. There are four pairs of adjacent sides.
  4. There are two pairs of opposite sides.
  5. Pairs of adjacent angles are $\angle\text{A}$, $\angle\text{B}$, $\angle\text{B}$, $\angle\text{C}$, $\angle\text{C}$, $\angle\text{D}$, $\angle\text{D}$, $\angle\text{A}$.
  6. Pairs of opposite angles are $\angle\text{A}$ and $\angle\text{C}$, $\angle\text{B}$ and $\angle\text{D}$.
  7. There are four pairs of adjacent angles.
  8. There are two pairs of opposite angles.
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