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Find the measure of the angle x in the given figure.
∠EFD+∠FED = x
(Exterior angle property of a triangle)
⇒ 28^{o}+42^{o }= ∠x or ∠x = 70^{o}
Which of the following options is INCORRECT?
∠1 =∠2 and ∠2 =∠3 [Alternate angles]
So, ∠1 =∠3 and ∠1+∠4+∠5 = 180^{o} [Angle sum property]
Also, ∠8 = ∠6
In the figure (not drawn to scale), ADF and BEF are triangles and EC = ED, find y.
In ΔCED, CE=ED
∴ ∠EDC = ∠ECD
[Angles opposite to equal sides are equal] ⇒ ∠ECD = 28^{o }
Also, ∠ECD = ∠BCA (Vertically opposite angles) ⇒ ∠BCA = 28^{o }
In ΔBCA, y = 62^{o}+28^{o}
Exterior angle property] ⇒ y = 90^{o}
In a ΔABC,which of the given condition holds?
In the figure (not drawn to scale), ABC is an equilateral triangle and ABD is an isosceles triangle with DA = DB, find x.
Since ABC is an equilateral triangle.
∴ ∠CAB =∠ABC =∠BCA = 60^{o }
And ∠DBA = ∠DAB = (60^{o}−x)
[∵ DA = DB]
In ΔDAB, ∠DAB∠DAB+∠ADB =180^{o }
⇒ 2(60^{o}−x)+88^{o }= 180^{o}
⇒ 2(60^{o}−x) = 92^{o}
⇒ 60^{o}−x = 46^{o }⇒ x = 14^{o}
ABC is an isosceles triangle with AB = AC and AD is altitude, then ____.
In the figure (not drawn to scale), ABCD is a square, ADE is an equilateral triangle and BFE is a straight line, find y.
In ΔAEB,
∠A=∠DAE+∠BAD ⇒ ∠A=60^{o}+90^{o}=15^{o }
And, AE=AB ⇒ ∠ABE=∠AEB
[Angles opposite to equal sides are equal]
Now, ∠A+∠ABE+∠AEB=180^{o } (Angle sum property)
⇒ 2∠AEB=180^{o }−150^{o }= 30^{o} ⇒ ∠AEB = 15^{o }
Now, ∠E=60^{o }⇒ ∠DEF=60^{o}−15^{o }= 45^{o }
∴ In ΔEFD, ∠DEF+∠EDF+∠EFD
= 180^{o }⇒ 45^{o}+60^{o}+y = 180^{o }
⇒ y = 180^{o}−(45^{o}+60^{o}) = 75^{o}
Find the measure of the angle x in the given figure.
∠UXV = y (Vertically opposite angles)
∴ y = 45^{o }In ΔXYZ y+x+63^{o} = 180^{o }
(Angle sum property)
⇒ 45^{o}+x+63^{o }= 180^{o}
⇒ x = 180^{o}−(45^{o}+63^{o})
⇒ x= 180^{o}−108^{o }= 72^{o}
The given figure shows three identical squares. Find x.
We have, ABCD, CEFG and CIHJ are all squares.
So, ∠1+∠2+x=90^{o }???
(i) 36^{o}+∠1+x=90^{o} q ??.
(ii) x+∠2+27^{o}=90^{o }??.
(iii) Adding (ii) and (iii),
we get 36^{o}+x+27^{o}+(∠A+∠2+x)=180^{o }
⇒ 63^{o}+x+90^{o}=180^{o }
(From (i)) ⇒ x=180^{o}−153^{o }= 27^{o}
In the figure (not drawn to scale), EFA is a rightangled triangle with ∠EFA=90^{o} and FGB is an equilateral triangle, find y−2x.
In ΔFGC, ∠CBF = 60^{o}
(Angle of equilateral triangle)
∴ x+60^{o}+92^{o }= 180^{o}
⇒ x = 180^{o}−152^{o}= 28^{o}
Now, In ΔBCF, ∠CBF = 60^{o}
∠FCB = 180^{o}−92^{o } (Linear pair)
⇒ ∠FCB=88^{o }
∴ ∠BFC+88^{o}+60^{o}=180^{o}
(Angle sum property)
⇒ ∠BFC=180^{o}−148^{o }= 32^{o}
And ∠AFE = 90^{o} ⇒ y+32^{o }= 90^{o}
⇒ y = 90^{o}−32^{o }= 58^{o}
∴ y−2x = 58^{o}−2×28^{o }= 58^{o}−56^{o }= 2^{o}
In the figure (not drawn to scale), ABC and DEF are two triangles, CA is parallel to FD and CFBE is a straight line. Find the value of x+y
∠FCA = ∠BFD
(Corresponding angles)
⇒ x = 51^{o}
Now, in ΔABC y = 51^{o}+83^{o} (Exterior angle property)
⇒ y = 134^{o }So, x+y = 51^{o}+134^{o }= 185^{o}
In a ΔABC,if AB+BC = 10 cm, BC+CA = 12 cm,CA+AB = 16 cm, then the perimeter of the triangle is ____.
It is given that,
AB+BC = 10cm ?..(i)
BC+CA = 12cm ?..(ii)
CA+AB = 16cm ?..(iii)
Adding (i), (ii) and (iii); we get
2(AB+BC+CA) = 10+12+16
⇒ AB+BC+CA = 19cm
In the figure, not drawn to scale, ACDF is a rectangle and BDE is a triangle. Find ∠BED
∠CDB+∠BDE = 90^{o }
(Angle of a rectangle)
⇒ 48^{o}+∠BDE = 90^{o}
⇒ ∠BDE = 90^{o}−48^{o }= 42^{o}
In ΔBED ∠EBD+∠BDE+∠BED = 180^{o}
(Angle sum property)
⇒ 86^{o}+42^{o}+∠BED = 180^{o }
⇒ ∠BED = 180^{o}−(86^{o}+42^{o}) = 52^{o}
In the figure, ABCD is a square, MDC is an equilateral triangle. Find the value of x.
∆MDC is an equilateral triangle. therefore all the angles will be 60°.
angle MCB=BCDMCD
=9060 = 30°
NOW,
In ∆BCD,
BC=CD
Therefore angle XBC = 45°.
when angle sum is applied in ∆CBX, X=105°
Find the measure of ∠LNM in the given figure.
∠KLO = ∠MLN
∴ ∠MLN = 70^{o} in ∠LMN,∠MLN+∠LNM+∠LMN = 180^{o }
(Angle sum property)
⇒ 70^{o}+∠LNM+50^{o}=180^{o
}⇒ ∠LNM = 180^{o}−(70^{o}+50^{o}) = 60^{o}
A 26 m long ladder reached a window 24 m from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.
In ΔPRQ, PR^{2 }= PQ^{2}+QR^{2 }
(By Pythagoras theorem)
(26)^{2 }=(24)^{2}+QR^{2 }or QR^{2 }
= 676−576 = 100
⇒ QR = √100 ⇒ QR = 10
∴ The distance of the foot of the ladder from the wall is 10 m.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Let KB is original height of the tree. In ΔABC,
AC^{2 }= AB^{2}+BC^{2 }= 52+122 =25+144=169
∴ AC = √169 = 13m
KB = KA+AB = (13+5)m = 18m
∴ Original height of the tree is 18m.
Aryan wants to plant a flower on the ground in the form of a rhombus. The diagonals of the rhombus measures 42 cm and 56 cm. Find the perimeter of the field.
Since diagonals of a rhombus bisect each other at 90^{o }.
Given: BD = 42 cm and AC = 56 cm
∴ BK = 1/2 BD = 42/2 = 21cm
AK=1/2 AC = 56/2 = 28 cm
In ΔKAB,AB^{2 }= AK^{2}+BK^{2 }= (28)^{2}+(21)^{2}=784+441=1225
∴ AB=√1225 = 35cm
∴ Perimeter of the field ABCD = 4×35 = 140 cm;
A 34 m long ladder reached a window 16 m from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.
Let AB = length of ladder, AC = height of window
In ΔABC, (AB)^{2 }= (AC)^{2}+(BC)^{2}
⇒ (34)^{2 }=(16)^{2}+BC^{2 }
or
BC^{2 }= (34)^{2}−(16)^{2 }⇒ BC^{2 }= 1156−256 = 900
∴ BC = √900 = 30m
Mrs Kaushik gives a problem to her students. Find the perimeter of a rectangle whose length is 28 cm and diagonal is 35 cm. What will be the correct answer?
ABCD is a rectangle,
In ΔACB, AC^{2 }= AB^{2}+BC^{2} (By Pythagoras theorem)
(35)^{2 }= (28)^{2}+BC^{2} or BC^{2 }= (35)^{2}−(28)^{2}
⇒ BC^{2 }= 1225−784 ⇒ BC^{2 }= 441
∴ BC==21cm
∴ Perimeter of rectangle
= 2×(28+21)cm = 2×(49)cm = 98cm
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