Q: ABC is a right triangle, right angled at C. If p is the length of perpendicular from C to AB & a, b, c have usual meaning, then prove that(i) pc = ab(ii) \frac{1}{{{p^2}}}\, = \,\,\frac{1}{{{a^2}}}\, + \,\frac{1}{{{b^2}}}

(i) Area of \Delta ABC = {1 \over 2}BC \times AC = {1 \over 2} \times a \times b = {{ab} \over 2} 3 area = {1 \over 2} \times CD \times AB {1 \over 2} \times p \times c = {{pc} \over 2} {{ab} \over 2} = {{pc} \over 2} ab = pc......(1) (ii) In \Delta ABC A{B^2} = A{C^2} + B{C^2} {\left( {{{ab} \over p}} \right)^2} = {b^2} + {a^2} {1 \over {{p^2}}} = {{{b^2} + {a^2}} \over {{a^2} + {b^2}}} {1 \over {{p^2}}} = {1 \over {{a^2}}} + {1 \over {{b^2}}}

Q: Find the diagonal of a square whose side is 10\ cm

The side of square is 10\ cm In right angled \triangle ABD AB^{2} + AD^{2} = BD^{2} [according to Pythagoras theorem] (10)^{2} + (10)^{2} = BD^{2} 100 + 100 = BD^{2} BD^{2} = 200 BD = 10\sqrt {2}cm \therefore Diagonal of square = 10 \sqrt {2}cm .

Question 1

Prove that the sum of angles of a triangle is 180.

Accepted Answer

Given :A triangle ABC.To prove : \angle A +\angle B +\angle C =180^{o} \implies \angle 1 +\angle 2 +\angle 3 =180^{o} Construction :Through A, draw a line l parallel to BC.Proof :Since  l \parallel BC . Therefore, \angle 2 =\angle 4  ....... eq(i) And,  \angle 3 =\angle 5 ...... eq(ii) adding    eq(i) and (ii) Therefore,  \angle 2 +\angle 3 =\angle 4 +\angle 5 \angle 1 + \angle 2 +\angle 3 =\angle 1 +\angle 4 +\angle 5    [adding \angle1 both Side] \angle 1 +\angle 2 +\angle 3 =180^{o} Thus, the sum of three angles of a triangle is  180^{o} .

Question 2

Prove that in a right triangle,the square of the hypotenuse is equal to the sum of the square of the other sides.

Accepted Answer

Given:- A right triangle  ABC  right angled at  B .To prove:-  {AC}^{2} = {AB}^{2} + {BC}^{2} Construction:- Draw  BD \perp AC Proof:-In  	riangle{ABC}  and  	riangle{ABD} \angle{ABC} = \angle{ADB} \quad \left( 	ext{Each } 90° ight) \angle{A} = \angle{A} \quad \left( 	ext{Common} ight) 	herefore 	riangle{ABC} \sim 	riangle{ABD} \quad \left( 	ext{By AA} ight) \cfrac{AB}{AC} = \cfrac{AD}{AB} \quad \left( \because 	ext{Sides of similar triangles are proportional} ight) \Rightarrow {AB}^{2} = AD \cdot AC ..... \left( 1 ight) Similarly, in  	riangle{ABC}  and  	riangle{BCD} \angle{ABC} = \angle{BDC} \quad \left( 	ext{Each } 90° ight) \angle{C} = \angle{C} \quad \left( 	ext{Common} ight) 	herefore 	riangle{ABC} \sim 	riangle{BCD} \quad \left( 	ext{By AA} ight) 	herefore \cfrac{DC}{BC} = \cfrac{BC}{AC} \Rightarrow {BC}^{2} = DC \cdot AC ..... \left( 2 ight) Adding equation  \left( 1 ight) \& \left( 2 ight) , we have {AB}^{2} + {BC}^{2} = AD \cdot AC + DC. AC \Rightarrow {AB}^{2} + {BC}^{2} = AC \left( AD + DC ight) \Rightarrow {AB}^{2} + {BC}^{2} = {AC}^{2} Hence proved.

Question 3

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.

Accepted Answer

Let A'CB represents the tree before it broken at the point C and let the top A' touches the ground at A after it broke. Then  \Delta ABC is a right angled triangle, right angled at B.AB =12 m and BC =5 mUsing Pythagoras theorem, In  \Delta ABC (AC)^2+(AB)^2+(BC)^2 \Rightarrow (AC)^2=(12)^2+(5)^2 \Rightarrow (AC)^2=144+25 \Rightarrow (AC)^2=169 \Rightarrow AC=13 mHence, the total height of the tree =AC+CB=13+5=18 m.

Question 4

ABC is a right triangle, right angled at C. If p is the length of perpendicular from C to AB & a, b, c have usual meaning, then prove that(i) pc = ab(ii)  \frac{1}{{{p^2}}}\, = \,\,\frac{1}{{{a^2}}}\, + \,\frac{1}{{{b^2}}}

Accepted Answer

(i) Area of  \Delta ABC = {1 \over 2}BC
	imes AC

= {1 \over 2} 	imes a 	imes b = {{ab}
\over 2} 3

area = {1 \over 2} 	imes CD 	imes AB

{1 \over 2} 	imes p 	imes c = {{pc}
\over 2}

{{ab} \over 2} = {{pc} \over 2}

ab = pc......(1)

(ii) In  \Delta ABC

A{B^2} = A{C^2} + B{C^2}

{\left( {{{ab} \over p}} ight)^2} =
{b^2} + {a^2}

{1 \over {{p^2}}} = {{{b^2} + {a^2}}
\over {{a^2} + {b^2}}}

{1 \over {{p^2}}} = {1 \over {{a^2}}} +
{1 \over {{b^2}}}

Question 5

In a  \Delta ABC, right angled at B,  AB = 24  cm,  BC = 7  cm. Determine(i)  \sin A, \cos A (ii)  \sin C, \cos C

Accepted Answer

R.E.F imageIn  \Delta   ABC, B is at right angle.Given, AB=24cmBC=7cmusing Pythagoras theorem  AB^{2}+BC^{2}=AC^{2} \Rightarrow (24)^{2}+(7)^{2}=AC^{2} \Rightarrow AC=\sqrt{(24^{2})+(7)^{2}}=\sqrt{576+49}=\sqrt{625} \Rightarrow AC=25 CM (i)sin A= \frac{BC}{AC}  = \frac{7}{25} cos A= \frac{AB}{AC} =\frac{24}{25} (ii)sin C= \frac{AB}{AC} =\frac{24}{25} cos c= \frac{BC}{AC} = \frac{7}{25}

Question 6

A  15 m long ladder reached a window  12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Accepted Answer

Let AC be the ladder and A be the window.Given:  AC=15 m,  AB=12 m,  CB=a mIn right angled triangle ACB, (Hypotenuse)^2=(Base)^2+[Perpendicular)^2  [By Pythagoras theorem] \Rightarrow (AC)^2=(CB)^2+(AB)^2 \Rightarrow (15)^2=(a)^2+(12)^2 \Rightarrow 225=a^2+144 \Rightarrow a^2=225-144=81 \Rightarrow a=\sqrt{81}=9 cmThus, the distance of the foot of the ladder from the wall is  9 m.

Question 7

In Figure, the side  QR  of  	riangle PQR  is produced to a point  S . If the bisectors of  \angle PQR  and  \angle PRS  meet at point  T , then prove that  \angle QTR=\dfrac { 1 }{ 2 } \angle QPR .

Accepted Answer

Given, Bisectors of  \angle PQR  and  \angle PRS  meet at point  T .To prove:  \angle QTR = \dfrac{1}{2}\angle QPR .Proof, \angle TRS = \angle TQR + \angle QTR  (Exterior angle of a triangle equals to the sum of the two interior angles.) \Rightarrow \angle QTR = \angle TRS - \angle TQR  --- (i)Also  \angle SRP =  \angle QPR + \angle PQR  2\angle TRS = \angle QPR + 2\angle TQR \angle QPR =  2\angle TRS - 2\angle TQR   \Rightarrow \dfrac{1}{2}\angle QPR =  \angle TRS - \angle TQR  --- (ii)Equating (i) and (ii), 	herefore \angle QTR  =\dfrac{1}{2}\angle QPR     [hence \, proved]

Question 8

Find the diagonal of a square whose side is  10\ cm

Accepted Answer

The side of square is  10\ cm In right angled  	riangle ABD AB^{2} + AD^{2} = BD^{2}  [according to Pythagoras theorem] (10)^{2} + (10)^{2} = BD^{2} 100 + 100 = BD^{2} BD^{2} = 200 BD = 10\sqrt {2}cm 	herefore  Diagonal of square  = 10 \sqrt {2}cm .

Question 9

The sides AB and AC of  \Delta ABC  are produced to points E and D respectively. If Bisectors BO and CO of,  \angle CBE\,\,and\,\,\angle BCD  respectively meet at point O, then prove that  \angle BOC = {90^0} - \frac{1}{2}\angle BAC.

Accepted Answer

∠CBE = 180 - ∠ABC∠CBO = 1/2 ∠CBE (BO is the bisector of ∠CBE)∠CBO = 1/2 ( 180 - ∠ABC)                                                   1/2 x 180 = 90 ∠CBO = 90 - 1/2 ∠ABC    .............(1)                                   1/2 x ∠ABC = 1/2∠ABC∠BCD = 180 - ∠ACD∠BCO = 1/2 ∠BCD     ( CO is the bisector os ∠BCD)∠BCO = 1/2 (180 - ∠ACD)∠BCO = 90 - 1/2∠ACD    .............(2)∠BOC = 180 - (∠CBO + ∠BCO)∠BOC = 180 - (90 - 1/2∠ABC + 90 - 1/2∠ACD)∠BOC = 180 - 180 + 1/2∠ABC + 1/2∠ACD∠BOC = 1/2 (∠ABC + ∠ACD)∠BOC = 1/2 ( 180 - ∠BAC)      (180 -∠BAC = ∠ABC + ∠ACD)∠BOC = 90 - 1/2∠BACHence proved

Question 10

In a triangle, if the square of one side is equal to the sum of the squares of other two sides, then the angle opposite to the first side is a right angle. Prove it

Accepted Answer

Given:-  ABC  is a triangle {AC}^{2} = {AB}^{2} + {BC}^{2} To prove:-  \angle{B} = 90° Construction:- Construct a triangle  PQR  right angled at  Q  such that,  PQ = AB  and  QR = BC Proof:-In  	riangle{PQR} {PR}^{2} = {PQ}^{2} + {QR}^{2} \quad \left( 	ext{By pythagoras theorem} ight) \Rightarrow {PR}^{2} = {AB}^{2} + {BC}^{2}..... \left( 1 ight) \quad \left( \because AB = PQ 	ext{ and } QR = BC ight) {AC}^{2} = {AB}^{2} + {BC}^{2} ..... \left( 2 ight) \quad \left( 	ext{Given} ight) From equation  \left( 1 ight) \& \left( 2 ight) , we have {AC}^{2} = {PR}^{2} \Rightarrow AC = PR ..... \left( 3 ight) Now, in  	riangle{ABC}  and  	riangle{PQR} AB = PQ BC = QR AC = PR \quad \left( 	ext{From } \left( 3 ight) ight) 	herefore 	riangle{ABC} \cong 	riangle{PQR} \quad \left( 	ext{By SSS congruency} ight) Therefore, by C.P.C.T., \angle{B} = \angle{Q} \because \angle{Q} = 90° 	herefore \angle{B} = 90° Hence proved.