Use the diagram to determine which statement is true.

The image illustrates the Pythagorean theorem, showing a right-angled triangle with squares constructed on each of its sides. The triangle is labeled as △ADG, with the right angle at vertex D. The squares on the legs AD and DG are colored green, while the square on the hypotenuse AG is also green. The grid background helps visualize the side lengths; AD is 3 units, DG is 4 units, and AG is 5 units. The squares on the sides correspond to the squares of these lengths:  Square on AD: 3x3 = 9 square units Square on DG: 4x4 = 16 square units Square on AG: 5x5 = 25 square units .CM8kHf text{fill:var(--m3c11)}.vM0jzc .CM8kHf text,.BFxDoe .CM8kHf text{fill:var(--m3c10)}.j86kh{display:inline-block;max-width:100%}ul.zVKf0d li.LLtSOc{list-style-type:none}.ibUR7b{display:flex;flex-direction:column}.LLtSOc,.ibUR7b{position:relative}.KEVENd{position:absolute;width:100%;height:100%;z-index:1}.igcdNb{position:absolute;width:100%;height:100%;background-color:var(--m3c6);opacity:50%;border-radius:16px}.mNme1d{-webkit-box-orient:vertical;-webkit-line-clamp:2;color:var(--m3c11);display:-webkit-box;font-family:Google Sans,Roboto,Arial,sans-serif;line-height:var(--m3t10);margin-bottom:4px;overflow:hidden}.ZigeC{color:var(--m3c10);display:flex;margin-bottom:8px;gap:16px}.y46mX{margin-right:4px}.gxZfx{flex:1;display:-webkit-box;-webkit-box-orient:vertical;overflow:hidden;-webkit-line-clamp:2}.LzPlYe{display:flex;align-items:baseline;gap:8px}.HAC9Fd{overflow:hidden}.tj8bcb{align-self:center;z-index:1}.tj8bcb .iTPLzd{width:20px;padding-left:0;padding-right:0}.tj8bcb .xTFaxe{top:unset}.LzPlYe .HAC9Fd.vDF3Oc .R8BTeb,.HAC9Fd .xuPcX.PCBdKc,.LzPlYe .xTFaxe{color:var(--m3c10)}.LW6JDf{margin-left:auto}.nfF3Me{min-height:36px}.buzUwc{min-height:29px}.iDBaYb{display:flex;position:relative;flex-wrap:nowrap;align-items:center}.LbKnXb{flex:1;min-width:0;width:calc(var(--google-fs,1)*100%)}.n6s7e.n6s7e{flex:none;flex-shrink:1;width:auto}.YAG2qc{margin-left:8px}.R8BTeb{font-size:14px;line-height:20px;align-items:center;letter-spacing:0.1px;color:var(--YLNNHc);order:0;align-self:stretch;flex-grow:0}.i0Rdmd{white-space:nowrap;overflow:hidden;text-overflow:ellipsis}.vDF3Oc.vDF3Oc .R8BTeb,.vDF3Oc.vDF3Oc .U3A9Ac{color:#ececec}.vDF3Oc.vDF3Oc .ob9lvb{color:#999da9;text-overflow:ellipsis;overflow:hidden}.J0EDnf{margin-left:12px}.Ss6Sxb{margin-left:0;margin-top:3px}.DMUiif.DMUiif{display:flex;align-items:center;white-space:nowrap;overflow:hidden;text-overflow:ellipsis}.B2BpC.B2BpC{color:var(--IXoxUe)}.du278d.du278d{font-size:12px;line-height:16px}.UYJxh.UYJxh{margin-left:6px}.iKOnjb.iKOnjb{margin-left:8px}.d6R6kf{align-self:center}.oRVWZ{border-radius:50%;display:flex;align-items:center;justify-content:center;width:fit-content}.CZ61cb{height:26px;width:26px;background-color:#fff;border:1px solid #525560}.qr5Pe{height:16px;width:16px;background-color:var(--TSWZIb)}.Hlkxk .bJVp8c{color:#5e5e5e}.g7Jr2c{height:28px;width:28px;background-color:#fff}.CZ61cb.GB3ilc,.GB3ilc{background-color:#fff}.PIcRzd{height:26px;width:26px;background-color:#fff;border:1px solid #3c4043}.mAl5Zc{border-radius:50%;content:"";height:26px;-webkit-mask:linear-gradient(#fff 0 0) content-box,linear-gradient(#fff 0 0);mask:linear-gradient(#fff 0 0) content-box,linear-gradient(#fff 0 0);-webkit-mask-composite:xor;mask-composite:exclude;padding:1px;position:absolute;width:26px}.Vwoesf:not(.oRVWZ){display:inline-block}.Vwoesf{vertical-align:middle}.XNo5Ab.XNo5Ab{display:block;}.lWlVCe{border-radius:50%}.bJVp8c{color:var(--TMYS9)}.Jj3Uob{color:#697988;color:#5e5e5e;}.KEVENd:hover:not(:has(.DgYKBe:hover))+.ibUR7b>.i3QiX>.W6DUrc>.mNme1d{text-decoration:underline}.KEVENd:hover+.ibUR7b>.mNme1d{text-decoration:underline}.IjabWd{margin-left:2px}.xTFaxe{top:3px}.xTFaxe{color:#9aa0a6}.D6lY4c{height:22px;width:22px;position:absolute;border-radius:11px}.iTPLzd{cursor:pointer;top:0;line-height:16px}.iTPLzd{left:0;width:28px}.iTPLzd{z-index:1}.iTPLzd.APQRre{height:18px;width:18px}.eY4mx{padding-left:12px}.lUn2nc{padding-right:12px} The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This can be written as \(a^{2}+b^{2}=c^{2}\), where \(c\) represents the length of the hypotenuse, and \(a\) and \(b\) represent the lengths of the other two sides. In this image: 
Responses
A Area(ABCD) - AREA(DGA) = AREA(DEFG)Area(ABCD) - AREA(DGA) = AREA(DEFG)
B Area(ABCD) - AREA(GHIA) = AREA(DGA)Area(ABCD) - AREA(GHIA) = AREA(DGA)
C Area(ABCD) + AREA(DGA) = AREA(GHIA)Area(ABCD) + AREA(DGA) = AREA(GHIA)
D Area(ABCD) + AREA(GHIA) = AREA(DEFG)Area(ABCD) + AREA(GHIA) = AREA(DEFG)
E Area(DEFG) + AREA(GHIA) = AREA(ABCD)

1 answer

Given the context of the Pythagorean theorem and the areas associated with the triangle \( \triangle ADG \) with sides of lengths 3, 4, and 5 units, we can analyze the statements provided in relation to the areas formed by the squares constructed on the triangle's sides.

The triangle has:

  • Side \( AD = 3 \) units, so the area of the square on \( AD \) is \( 3^2 = 9 \) square units.
  • Side \( DG = 4 \) units, so the area of the square on \( DG \) is \( 4^2 = 16 \) square units.
  • Hypotenuse \( AG = 5 \) units, so the area of the square on \( AG \) is \( 5^2 = 25 \) square units.

According to the Pythagorean theorem: \[ AD^2 + DG^2 = AG^2 \] which translates into: \[ 9 + 16 = 25 \] This equation confirms the relationship \( \text{Area of square on } AD + \text{Area of square on } DG = \text{Area of square on } AG \).

Now, let's evaluate the statements based on these area calculations:

  • A: Area(ABCD) - AREA(DGA) = AREA(DEFG)
  • B: Area(ABCD) - AREA(GHIA) = AREA(DGA)
  • C: Area(ABCD) + AREA(DGA) = AREA(GHIA)
  • D: Area(ABCD) + AREA(GHIA) = AREA(DEFG)
  • E: Area(DEFG) + AREA(GHIA) = AREA(ABCD)

From what we established:

  1. The area of the square on \( AD \) (9) and the area of the square on \( DG \) (16) summing to the area of the square on \( AG \) (25) can imply specific relationships regarding the areas if we consider comparable areas.

Given that:

  • C corresponds directly to the Pythagorean theorem \( \text{Area of square on } AD + \text{Area of square on } DG = \text{Area of square on } AG \), which can be denoted as \( 9 + 16 = 25 \).

Based on the above analysis, statement C: \( \text{Area(ABCD)} + \text{AREA(DGA)} = \text{AREA(GHIA)} \) is true as it aligns directly with the Pythagorean theorem since we verify the relationship between the areas related to the three sides.

Thus, the correct answer is: C: Area(ABCD) + AREA(DGA) = AREA(GHIA)