आलोक पुंज माध्यमिक विद्यालय (1104159)
करावल नगर, दिल्ली-94
परीक्षा
कक्षा 7
विषय गणित पूर्णाक 50
समय 2 घंटे
नोट सभी प्रश्न अनिवार्य हैं।
प्रश्न 1 रिक्त स्थानों को पूरा करो
2x10%3D20
0 (-7)+11
(iv) 5%
(v) 10x10x10% (vi) % +% = (vii)0.50+0.40 = -
(ix) 2x2x2x2 =
(x) त्रिमुज के तीनों कोंणों का योग
प्रश्न 2 सत्य/असत्य बताओ।
1x10 = 10
(ii) -5x-4=_
(III) 10°
(viii)9%
GO
BOUW
होता है।
(i) 25
(il) (-1) (-1) = 1
(ii) वर्ग में सममित रेखा चार हैं
(iv) वर्ग का परिमाप
%3D4x भुजा
(v) त्रिभुज का क्षेत्रफल =Xx आधार x उचाई
(vi)-7 छोटा है 2 से
(vil) 0>-10
(vill) * परिमेय संख्या है
(ix) आयत का परिमाप = 2 (लम्बाई+चौड़ाई)
(x) 70° का पूरक 20 है
प्रश्न 3
2x5-10
(1) आयत एंव वर्ग की आकृति बनाओ एंव इनके परिमाप का सूत्र लिखिए।
P.T.O.
Answers
Answer:
Answer:
\begin{gathered} \\ \end{gathered}
\begin{gathered}\frak{Given}\begin{cases}\sf{Length\:of\:the\:wire=\bf 44\:cm.}\\\sf{First,it\:is\:bent\:to\:form\:a\:\bf Circle. }\\\sf{Then,it\:is\:re-bent\:to\:form\:a\:\bf Square. }\end{cases}\end{gathered}
Given
⎩
⎪
⎪
⎨
⎪
⎪
⎧
Lengthofthewire=44cm.
First,itisbenttoformaCircle.
Then,itisre−benttoformaSquare.
\begin{gathered} \\ \end{gathered}
\underline{\bf\bigstar\:To\:Find:-}
★ToFind:−
\begin{gathered} \\ \end{gathered}
Radius of the circle formed.
Area of the circle.
Length of each side of the square formed.
The figure (square/circle) which encloses more area
\begin{gathered} \\ \end{gathered}
\underbrace{\underline{\bf\bigstar\:Solution:-}}
★Solution:−
\begin{gathered} \\ \end{gathered}
» Since the same wire is used throughout so, the perimeter of every figure formed would be equal to the length of the wire.
\begin{gathered} \\ \end{gathered}
◆ Finding the Radius and the Area of the Circle :
\begin{gathered} \\ \end{gathered}
» The length of the wire would be equal to the circumference of the circle.
So,
\begin{gathered} \colon \rarr \sf \: length _{ \: wire} = circumference_ { \: circle} \\ \\ \colon \rarr \bf \: \cancel{44 }\: {cm }= \bf \cancel 2\pi r \\ \\ \displaystyle\colon \rarr \bf \: \cancel{22} \: cm = \dfrac{ \cancel{22}}{7} \times r \\ \\ \colon \dashrightarrow \boxed{\boxed{\pink{ \frak{ r = 7 \: cm}}}}\end{gathered}
:→length
wire
=circumference
circle
:→
44
cm=
2
πr
:→
22
cm=
7
22
×r
:⇢
r=7cm
And,
\begin{gathered} \colon \rarr \sf \: area _{ \: circle} = \pi {r}^{2} \\ \\ \colon \bf \rarr \: area _{ \: circle}= \frac{22}{ \cancel7} \times \cancel 7 \times 7 \: {cm}^{2} \\ \\ \colon \rarr \bf \: area _{ \: circle} = 22 \times 7 \: {cm}^{2} \\ \\ \colon \dashrightarrow \boxed{ \boxed{ \purple{ \frak{area _{ \: circle} = 154 \: {cm}^{2} }}}}\end{gathered}
:→area
circle
=πr
2
:→area
circle
=
7
22
×
7
×7cm
2
:→area
circle
=22×7cm
2
:⇢
area
circle
=154cm
2
\begin{gathered} \\ \end{gathered}
◆ Finding the side and the area of the square :
\begin{gathered} \\ \end{gathered}
» The length of the wire would be equal to the perimeter of the square.
So,
\begin{gathered} \colon \rarr \sf \: length _{ \: wire} = perimeter _{ \: square} \\ \\ \colon \rarr \bf \: \cancel{44 }\: cm = \cancel4 \times side \\ \\ \colon \dashrightarrow \boxed{ \boxed{ \frak{ \pink{side = 11 \: cm}}}}\end{gathered}
:→length
wire
=perimeter
square
:→
44
cm=
4
×side
:⇢
side=11cm
And,
\begin{gathered} \colon \sf \rarr \: area _{ \: square} = {side}^{2} \\ \\ \colon \rarr \bf \: area _{ \: square} = {(11 \: cm)}^{2} \\ \\ \colon \dashrightarrow \boxed{ \boxed{ \frak{ \pink{area _{ \: square} = 121 \: {cm}^{2} }}}}\end{gathered}
:→area
square
=side
2
:→area
square
=(11cm)
2
:⇢
area
square
=121cm
2
\begin{gathered} \\ \end{gathered}
◆ Figure which encloses more area :
\begin{gathered} \\ \end{gathered}
» We know
Area of Circle = 154 cm²
Area of Square = 121 cm²
» The value of Area of Circle is greater than the value of the Area of Square .
\begin{gathered} \\ \end{gathered}
\therefore\;\underline{\bf \red{Circle}\:\sf{encloses\:more\:area.}}∴
Circleenclosesmorearea.
\begin{gathered} \\ \end{gathered}
_____________________________