ᴘʟᴇᴀsᴇ sᴏʟᴠᴇ ɪᴛ ғᴀsᴛ.
ᴅᴏɴ'ᴛ ᴅᴀʀᴇ ᴛᴏ sᴘᴀᴍ.
Answers
Q.1.
PQRS is a quadrilateral in which diagonal PR and QS intersect at a O .
to prove - PQ +QR +RS+SP < 2 ( PR + QS )
proof -
we know that sum of any two side of a triangle is greater than the third side .
.'. in Δ PQO ,
PO+QO>PQ , .......................(i)
in Δ SOP
SO + PO >PS , .........................(ii)
in Δ SOR
SO + OR > RS ...........................(iii)
in Δ QOR ,
QO + OR > QR ...........................(iv)
on adding eqn. i , ii , iii & iv
we get ,
PO+QO+SO+PO+SO+OR+QO+OR > PQ+PS+SR+QR
also ⇒ 2 ( PO + QO + SO + OR ) > PQ+PS+SR + QR
= 2( PR + QS ) > PQ+PS+RS + QR ( proved) ....
Q.2.
(i)The sum of 3 times n and seven is 1.
(ii)subtracting 5 from x gives 8
(iii)subtracting 3 from 1/5 of n gives 6
Q.3.
The first thing we need to do is distribute the -2(1-x) we do this by multiplying the -2 by both 1 and -x.
-2*1=-2
-2*-x=2x
Now we can put them in our problem:
3-5x+2x=-2-2+2x
We are going to put our x's on the left side now:
3-5x+2x-2x Remember, when the x's or numbers cross the equals sign, change the sign!
Now we are going to put the numbers in the right side:
-5x+2x-2x=-3-2-2
Combine like terms:
-5x=-7
Finally, divide both sides by -5
x= -7/-5
Remember, if both numbers in a fraction or negative, it is a positive fraction
Final Answer: [x= 7/5]
Q.4.
(i)
triangle APB
angle sum property = 180
angle A + Angle P + z=180
100+50+z=180
150+z=180
z=30
100+y = 180 (straight line is 180 )
y =80
TRIANGLE ABC
ANGLE A +ANGLE B+ ANGLE C =180 (ANGLE SUM PROPERTY)
Y+X+75 =180
80+X +75 =180
155+X =180
X=180-155
X=25
X=25
Y=80
Z=30
(ii)
angle BAC = angle FAE (vertically opp. angle)
angle BAC = x = 70°
angle BCA + angle ACD = 180° (linear pair)
z + 110 = 180°
angle BCA = z = 70°
angle BCA + angle ABC + angle BAC = 180° ( sum of angles of a triangle is equal to 180°)
z + y + x = 180°
70° +70° +y = 180°
angle ABC = y = 40 °
x = 70°
y = 40°
z = 70°
Answer:
Q.1.
PQRS is a quadrilateral in which diagonal PR and QS intersect at a O .
to prove - PQ +QR +RS+SP < 2 ( PR + QS )
proof -
we know that sum of any two side of a triangle is greater than the third side .
.'. in Δ PQO ,
PO+QO>PQ , .......................(i)
in Δ SOP
SO + PO >PS , .........................(ii)
in Δ SOR
SO + OR > RS ...........................(iii)
in Δ QOR ,
QO + OR > QR ...........................(iv)
on adding eqn. i , ii , iii & iv
we get ,
PO+QO+SO+PO+SO+OR+QO+OR > PQ+PS+SR+QR
also ⇒ 2 ( PO + QO + SO + OR ) > PQ+PS+SR + QR
= 2( PR + QS ) > PQ+PS+RS + QR ( proved) ....
Q.2.
(i)The sum of 3 times n and seven is 1.
(ii)subtracting 5 from x gives 8
(iii)subtracting 3 from 1/5 of n gives 6
Q.3.
The first thing we need to do is distribute the -2(1-x) we do this by multiplying the -2 by both 1 and -x.
-2*1=-2
-2*-x=2x
Now we can put them in our problem:
3-5x+2x=-2-2+2x
We are going to put our x's on the left side now:
3-5x+2x-2x Remember, when the x's or numbers cross the equals sign, change the sign!
Now we are going to put the numbers in the right side:
-5x+2x-2x=-3-2-2
Combine like terms:
-5x=-7
Finally, divide both sides by -5
x= -7/-5
Remember, if both numbers in a fraction or negative, it is a positive fraction
Final Answer: [x= 7/5]
Q.4.
(i)
triangle APB
angle sum property = 180
angle A + Angle P + z=180
100+50+z=180
150+z=180
z=30
100+y = 180 (straight line is 180 )
y =80
TRIANGLE ABC
ANGLE A +ANGLE B+ ANGLE C =180 (ANGLE SUM PROPERTY)
Y+X+75 =180
80+X +75 =180
155+X =180
X=180-155
X=25
X=25
Y=80
Z=30
(ii)
angle BAC = angle FAE (vertically opp. angle)
angle BAC = x = 70°
angle BCA + angle ACD = 180° (linear pair)
z + 110 = 180°
angle BCA = z = 70°
angle BCA + angle ABC + angle BAC = 180° ( sum of angles of a triangle is equal to 180°)
z + y + x = 180°
70° +70° +y = 180°
angle ABC = y = 40 °
x = 70°
y = 40°
z = 70°
thnx for points